Atomic Foundations of Matter — Notes

Law of Conservation of Mass

The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. This means that the total mass of the reactants before a chemical reaction is always equal to the total mass of the products after the reaction.

This fundamental principle was proposed by Antoine Lavoisier in 1789 and is often referred to as the Father of Modern Chemistry. Lavoisier's observation was that in any process, the total amount of matter remains constant.

How it Works and Why it Matters

Chemical reactions involve the rearrangement of atoms, but the atoms themselves are not lost or gained. Imagine a set of building blocks: you can rearrange them into many different structures, but the total number and mass of the blocks remain the same. Similarly, in a chemical reaction, the atoms present in the reactants combine in new ways to form the products, but the total mass of these atoms is conserved.

Example: When hydrogen gas ($H_2$) reacts with oxygen gas ($O_2$) to form water ($H_{2}O$), the reaction is represented as:

If we take 4 grams of hydrogen and react it completely with 32 grams of oxygen, we will always get 36 grams of water.

• Total mass of reactants: 4 g ($H_2$) + 32 g ($O_2$) = 36 g
• Total mass of products: 36 g ($H_{2}O$)

The mass is conserved because the atoms of hydrogen and oxygen are simply rearranged to form water molecules.

Experimental Verification

Activities in the lab help demonstrate this law:

• Activity with Sodium Sulfate and Barium Chloride: When solutions of sodium sulfate ($N{a}_{2}S{O}_4$) and barium chloride ($BaC{l}_2$) are mixed in a closed system, a precipitate of barium sulfate ($BaS{O}_4$) and sodium chloride ($NaCl$) are formed.

If the mass of the reactants is measured before mixing and the mass of the products is measured after mixing using a precise weighing balance, the total mass will remain the same, even though a solid precipitate has formed and dissolved ions are now present.

• Reaction in an Open System: If a gas is produced in an open container (like burning something in the air), the gas can escape, leading to a measured decrease in mass. This does not violate the law; it simply means that the mass of the escaping gas was not included in the final measurement. The law holds true if all reactants and products are accounted for, often requiring a closed system for gaseous reactions.

Application in Calculations

The Law of Conservation of Mass is crucial for stoichiometric calculations, where we predict the amount of products formed from given amounts of reactants.

Example: If 2.4 g of carbon reacts completely with oxygen, how much carbon dioxide will be produced?

We know from the reaction:

That 12 g of carbon (atomic mass) reacts with 32 g of oxygen to produce 44 g of carbon dioxide.

Using the Law of Conservation of Mass (mass of reactants = mass of products): Mass of carbon + Mass of oxygen = Mass of carbon dioxide 12 g + 32 g = 44 g

Now, if 2.4 g of carbon reacts: We can set up a proportion or use the ratio from the balanced equation. The ratio of carbon to carbon dioxide is 12:44. So, for 2.4 g of carbon:

Therefore, 8.8 g of carbon dioxide will be produced.

Remember

The Law of Conservation of Mass applies to all chemical reactions and physical changes, provided no matter enters or leaves the system being considered.

Exam-smart

In case or source-based questions, you might be given the masses of reactants and products in a chemical reaction. You will need to apply the Law of Conservation of Mass to verify if the law is obeyed by comparing the total mass of reactants and products, or to calculate the mass of an unknown reactant or product. Be mindful of whether the reaction is described as occurring in an open or closed system, especially if gases are involved.

Law of Constant Proportions

The Law of Constant Proportions, also known as the Law of Definite Proportions or Proust's Law, is a fundamental principle in chemistry. It states that a chemical compound always consists of the same elements combined together in the same definite proportion by mass, regardless of its source or method of preparation.

This means that if you have a pure sample of a specific compound, the ratio of the masses of its constituent elements will always be the same.

How it Works: An Example with Water

Let's consider water ($H_{2}O$).

• Water can be collected from various sources: rivers, wells, oceans, or even synthesized in a laboratory.
• According to the Law of Constant Proportions, no matter where the water comes from, if you decompose it, you will always find that it is made up of hydrogen and oxygen in a fixed mass ratio.
• This fixed ratio for water is 1:8. For every 1 gram of hydrogen, there are always 8 grams of oxygen.
• This proportion can be represented as:

Why this is important: This law tells us that compounds are not just random mixtures of elements. They have a specific, fixed composition. This fixed composition is a direct consequence of the atomic nature of matter, where elements combine in whole-number ratios of atoms.

Example: Sodium Chloride (NaCl)

Sodium chloride, commonly known as table salt, also adheres to this law.

• It contains sodium (Na) and chlorine (Cl).
• The mass ratio of sodium to chlorine in NaCl is 23:35.5.

• Problem: If 46 g of sodium reacts completely to form NaCl, how much chlorine is needed?
• Solution: We know the ratio is 23 g Na to 35.5 g Cl. If we have 46 g of sodium (which is 2 × 23 g), we will need 2 × 35.5 g of chlorine. Mass of chlorine required = $(35.5÷23)\times 46=71$ g.

Distinction Between Compounds and Mixtures

The Law of Constant Proportions is a key differentiator between compounds and mixtures.

• Compounds have a fixed composition by mass.
• Mixtures have a variable composition by mass.

For instance, air is a mixture of gases like nitrogen ($N_2$) and oxygen ($O_2$). The proportion of these gases can vary depending on location and altitude. However, water ($H_{2}O$) will always have hydrogen and oxygen in a 1:8 mass ratio.

Application and Reasoning: Copper Carbonate

• Students X and Y prepared an oxide of copper. Student X used a copper to oxygen ratio of 4:1 by mass. Student Y used a ratio of 8:2 by mass.
• Analysis of Student Y's ratio: The ratio 8:2 simplifies to 4:1 (by dividing both numbers by 2).
• Conclusion: Both students used the same proportion of copper to oxygen. Therefore, if they successfully formed the same copper compound, their results justify the Law of Constant Proportions. The law holds true because it emphasizes the ratio (which can be simplified), not the absolute masses used.

Remember

The Law of Constant Proportions applies to the elements within a pure compound. The ratio is by mass, and it remains constant regardless of the amounts of reactants used or the source of the compound.

Exam-smart

You might be given the mass of one element and the ratio of elements in a compound and asked to calculate the mass of the other element. Alternatively, you might be given different preparation methods or sources for a compound and asked to determine if they are consistent with this law. The key is to simplify any given ratios to their lowest terms to check for constancy.

Dalton’s Atomic Theory

John Dalton proposed his atomic theory in 1808, which laid the groundwork for modern atomic understanding. His theory consists of several key postulates (fundamental assumptions accepted without proof):

• Postulate 1: All matter is composed of extremely small, indivisible particles called atoms.
• These atoms are the fundamental building blocks of all substances.
• Postulate 2: Atoms cannot be created or destroyed in a chemical reaction.
• This explains the Law of Conservation of Mass: the total mass of reactants equals the total mass of products because atoms are only rearranged, not lost or gained.
• Postulate 3: Atoms of a given element are identical in mass and chemical properties.
• All atoms of, say, oxygen, are the same as each other.
• Postulate 4: Atoms of different elements have different masses and chemical properties.
• An oxygen atom is fundamentally different from a carbon atom.
• Postulate 5: Atoms combine in the ratio of simple whole numbers to form compounds.
• This explains the Law of Constant Proportions: a chemical compound always contains the same elements in the same proportion by mass, regardless of its source. For example, water ($H_{2}O$) always consists of two hydrogen atoms and one oxygen atom.
• Postulate 6: The relative number and kinds of atoms are constant in a given compound.
• This reiterates that the composition of a specific compound is fixed.

Example: When magnesium burns in air, it forms magnesium oxide. This illustrates Dalton's theory as magnesium atoms combine with oxygen atoms to form magnesium oxide. The atoms themselves are not changed; they are simply rearranged into a new compound.

Remember

Dalton's theory provided the first scientific explanation for the laws of chemical combination (Conservation of Mass and Constant Proportions).

Exam-smart

Questions might present a scenario (e.g., burning fuel) and ask which postulate of Dalton's theory explains why mass is conserved. You would relate it to atoms being indivisible and merely rearranged.

How Atoms Combine?

Atoms are the fundamental units of elements. They can exist independently or combine with other atoms to form larger structures.

• Molecules of Elements: Two or more atoms of the same element can join together to form a molecule of that element. For example, a hydrogen molecule ($H_2$) is formed from two hydrogen atoms.
• Molecules of Compounds: Atoms of different elements can join together to form a molecule of a compound. For example, a molecule of hydrogen chloride (HCl) is formed from one hydrogen atom and one chlorine atom.

A molecule is defined as an electrically neutral group of two or more atoms held together by chemical bonds. It is the smallest particle of a substance that can exist independently and retain the chemical properties of that substance.

Some elements, like helium (He), exist only as individual atoms because their atoms are already stable (have a full outermost electron shell).

Atoms achieve stability by having a complete outermost electron shell (usually 8 electrons, known as an octet, or 2 electrons in the first shell, a duplet). If an atom's outermost shell is not full, it can become stable by:

1. Sharing its valence electrons with another atom.
2. Transferring (gaining or losing) its valence electrons to or from another atom.

The force that holds atoms together in molecules and compounds is called a chemical bond. The process of atoms combining leads to a more stable state with lower energy.

Bonding by sharing of electrons — Covalent Bond

A covalent bond is formed when two atoms share one or more pairs of valence electrons to achieve a stable electronic configuration. This sharing typically occurs between non-metal atoms.

• Molecules of Elements:

• Hydrogen Molecule ($H_2$): Each hydrogen atom has 1 electron. To achieve a stable duplet (2 electrons), each atom shares its electron with the other. This creates one shared pair of electrons, forming a single covalent bond (H—H).

• Chlorine Molecule ($C{l}_2$): Each chlorine atom has 7 valence electrons. It needs 1 more electron to complete its octet. Two chlorine atoms share one electron each, forming one shared pair and a single covalent bond (Cl—Cl).

• Oxygen Molecule ($O_2$): Each oxygen atom has 6 valence electrons. It needs 2 more electrons. Two oxygen atoms share two electrons each, forming two shared pairs and a double covalent bond (O=O).

• Molecules of Compounds:

• Hydrogen Chloride (HCl): Hydrogen needs 1 electron, and chlorine needs 1 electron. They share one electron pair, forming a single covalent bond (H—Cl).

• Water ($H_{2}O$): Oxygen needs 2 electrons, and each hydrogen needs 1 electron. The oxygen atom shares one electron with each of the two hydrogen atoms. This results in two single covalent bonds (H—O—H).

Remember

Covalent bonds form between atoms that need to share electrons to achieve a stable electron configuration. The number of shared electron pairs determines if it's a single, double, or triple bond.

Exam-smart

You might be given the electronic configuration of elements and asked to draw the formation of a molecule (like $H_2$, $O_2$, $H_{2}O$, $C{O}_2$) showing the shared electron pairs and the resulting covalent bond.

Bonding by electron transfer — Ionic bond

An ionic bond is formed by the complete transfer of one or more valence electrons from one atom to another. This typically occurs between a metal (which tends to lose electrons) and a non-metal (which tends to gain electrons).

1. Formation of Ions:

• Cations: Atoms that lose electrons become positively charged and are called cations. For example, a sodium atom (Na) with 1 valence electron loses it to become a sodium ion ($N{a}^{+}$) with a stable electron configuration.
• Anions: Atoms that gain electrons become negatively charged and are called anions. For example, a chlorine atom (Cl) with 7 valence electrons gains 1 electron to become a chloride ion ($C{l}^{-}$) with a stable octet.

2. Electrostatic Attraction: The positively charged cations and negatively charged anions are held together by a strong electrostatic force of attraction. This force is the ionic bond.

Example: Sodium Chloride (NaCl)

• Sodium (Na) has 1 valence electron. It loses this electron to become $N{a}^{+}$.
• Chlorine (Cl) has 7 valence electrons. It gains 1 electron to become $C{l}^{-}$.
• The $N{a}^{+}$ and $C{l}^{-}$ ions are then strongly attracted to each other, forming the ionic compound sodium chloride.

Ionic compounds often form crystal lattices, which are three-dimensional arrangements of ions held together by ionic bonds. In NaCl, each $N{a}^{+}$ ion is surrounded by $C{l}^{-}$ ions, and vice-versa.

Remember

Ionic bonds are formed by the electrostatic attraction between oppositely charged ions created by electron transfer. Metals form cations, and non-metals form anions.

Exam-smart

You might be asked to explain the formation of an ionic compound like magnesium chloride ($MgC{l}_2$). You would need to show the loss of electrons from magnesium to form $M{g}^{2+}$ and the gain of electrons by two chlorine atoms to form two $C{l}^{-}$ ions, followed by their electrostatic attraction.

Naming Ionic Compounds

Ionic compounds are formed by the electrostatic attraction between positively charged ions (cations) and negatively charged ions (anions). When naming these compounds, the cation is always written first, followed by the anion.

• Simple anions (ions formed from a single atom) typically have names ending in -ide. For example, $C{l}^{-}$ is chloride, $O^{2-}$ is oxide, and $S^{2-}$ is sulfide.
• Metals usually form cations, and non-metals usually form anions. Therefore, ionic compounds are commonly formed between metals and non-metals.
• Some ions are formed from a group of atoms bonded together, carrying an overall charge. These are called polyatomic ions. Their names usually do not end in -ide. For example, $O{H}^{-}$ is hydroxide, $N{O}_3^{-}$ is nitrate, and $C{O}_3^{2-}$ is carbonate.

Table 9.1: Common Ions

Writing Chemical Formulae

Chemical formulae represent the elemental composition of a compound. They show the types of atoms and their relative numbers in a molecule or formula unit.

Writing chemical formulae of covalent compounds

Covalent compounds are formed by sharing of electrons between non-metal atoms. The formula indicates the exact number of atoms of each element in a molecule.

1. Identify the symbols of the elements involved.
2. Determine the valency of each element. Valency is the combining capacity of an element, often equal to the number of electrons an atom can gain, lose, or share to achieve a stable electron configuration.
3. Criss-cross the valencies: Write the valency of one element as a subscript for the other element. This indicates how many atoms of each element are needed to form a neutral molecule.
4. If the valency is 1, it is not written as a subscript.

Example: Hydrogen Chloride

• Symbol of Hydrogen (H) and Chlorine (Cl)
• Valency of H is 1, Valency of Cl is 1.
• Criss-crossing gives $H_{1}C{l}_1$.
• Since the valency is 1, it is written as HCl.

Example: Carbon Tetrachloride

• Symbol of Carbon (C) and Chlorine (Cl)
• Valency of C is 4, Valency of Cl is 1.
• Criss-crossing gives $C_{1}C{l}_4$.
• The formula is $CC{l}_4$.

Writing chemical formulae of ionic compounds

Ionic compounds are formed by the transfer of electrons, resulting in ions. The chemical formula represents the simplest whole-number ratio of ions in the compound, ensuring the overall electrical neutrality.

1. Write the symbols of the cation first, followed by the anion.
2. Write the charges of the ions below their respective symbols (e.g., $C{a}^{2+}$ becomes $Ca2+$).
3. Criss-cross the charges: Transfer the numerical value of the charge from one ion to become the subscript of the other ion. This ensures that the total positive charge equals the total negative charge.
4. Simplify the subscripts: If the subscripts have a common factor, divide them by the greatest common divisor to get the simplest whole-number ratio.
5. When dealing with polyatomic ions, if more than one polyatomic ion is needed to balance the charges, enclose the polyatomic ion in parentheses () and place the subscript outside the parentheses.

Example: Calcium Chloride

• Cation: Calcium ($C{a}^{2+}$), Anion: Chloride ($C{l}^{-}$)
• Symbols and charges: $Ca2+$ and $Cl1-$
• Criss-crossing charges: $Ca$ gets a subscript of 1, $Cl$ gets a subscript of 2.
• Formula: $CaC{l}_2$. (One calcium ion for two chloride ions).

Example: Aluminium Oxide

• Cation: Aluminium ($A{l}^{3+}$), Anion: Oxide ($O^{2-}$)
• Symbols and charges: $Al3+$ and $O2-$
• Criss-crossing charges: $Al$ gets a subscript of 2, $O$ gets a subscript of 3.
• Formula: $A{l}_2{O}_3$. (Two aluminium ions for three oxide ions).

Example: Magnesium Oxide

• Cation: Magnesium ($M{g}^{2+}$), Anion: Oxide ($O^{2-}$)
• Symbols and charges: $Mg2+$ and $O2-$
• Criss-crossing charges: $Mg$ gets a subscript of 2, $O$ gets a subscript of 2.
• Simplified formula: MgO. (The ratio 2:2 simplifies to 1:1).

Example: Magnesium Hydroxide

• Cation: Magnesium ($M{g}^{2+}$), Anion: Hydroxide ($O{H}^{-}$)
• Symbols and charges: $Mg2+$ and $OH1-$
• Criss-crossing charges: $Mg$ gets a subscript of 1, $OH$ gets a subscript of 2.
• Formula: $Mg(OH{)}_2$. (Brackets are used because we need two hydroxide ions).

Example: Aluminium Sulfate

• Cation: Aluminium ($A{l}^{3+}$), Anion: Sulfate ($S{O}_4^{2-}$)
• Symbols and charges: $Al3+$ and $S{O}_{4}2-$
• Criss-crossing charges: $Al$ gets a subscript of 2, $S{O}_4$ gets a subscript of 3.
• Formula: $A{l}_2(S{O}_4{)}_3$. (Brackets are used for the polyatomic sulfate ion).

Remember

The chemical formula of an ionic compound represents the simplest whole-number ratio of ions that results in a neutral compound.

Exam-smart

You might be given the names of ions or their charges and asked to write the chemical formula, or vice versa. You may also see questions describing the formation of compounds from elements and asking for the formula and type of bond.

Properties of the Ionic and the Covalent Compounds

Ionic and covalent compounds exhibit distinct physical properties due to their different bonding structures.

Reasoning for Properties:

• Solubility: Ionic compounds often dissolve in polar solvents like water because the polar water molecules can surround and separate the ions. Non-polar solvents like kerosene do not interact effectively with ions. Covalent compounds, being generally non-polar, dissolve better in non-polar solvents.
• Electrical Conductivity: In the solid state, ions in ionic compounds are held rigidly in a crystal lattice and cannot move to carry charge. When melted or dissolved in water, these ions become mobile and can conduct electricity. Covalent compounds, whether solid, liquid, or gas, typically do not have free-moving charged particles to conduct electricity.
• Melting/Boiling Points: The strong electrostatic forces of attraction between oppositely charged ions in ionic compounds require a large amount of energy to overcome, leading to high melting and boiling points. Covalent bonds within molecules are strong, but the forces between molecules (intermolecular forces) are often weak, resulting in lower melting and boiling points.

Remember

The key difference in electrical conductivity is the presence of mobile charged particles. Ionic compounds develop this mobility when molten or dissolved; covalent compounds generally lack it entirely.

Exam-smart

You might be presented with data from experiments on solubility and conductivity and asked to classify a substance as ionic or covalent, or predict its properties based on its type. For instance, a substance insoluble in water but soluble in petrol, and a poor conductor of electricity in all states, is likely covalent.

Molecular Mass of Covalent Compounds

Molecular mass is the sum of the atomic masses of all the atoms present in one molecule of a substance. This concept applies specifically to covalent compounds because they exist as discrete molecules.

To calculate the molecular mass of a covalent compound:

1. Identify the chemical formula of the compound. This formula tells us the number of atoms of each element in one molecule.
2. Find the atomic mass of each element from the periodic table. Atomic mass is usually expressed in atomic mass units (u).
3. Multiply the atomic mass of each element by the number of atoms of that element present in the molecule.
4. Add up the values obtained in step 3 for all the elements in the molecule. The total sum is the molecular mass.

Example 1: Molecular Mass of Water ($H_{2}O$)

• The chemical formula is $H_{2}O$. This means one molecule of water contains 2 hydrogen atoms and 1 oxygen atom.
• Atomic mass of Hydrogen (H) = 1 u
• Atomic mass of Oxygen (O) = 16 u

Molecular mass of $H_{2}O$ = (Atomic mass of H × Number of H atoms) + (Atomic mass of O × Number of O atoms) Molecular mass of $H_{2}O$ = ($1\text{ u}\times 2$) + ($16\text{ u}\times 1$) Molecular mass of $H_{2}O$ = $2\text{ u}+16\text{ u}$ Molecular mass of $H_{2}O$ = 18 u

Example 2: Molecular Mass of Carbon Dioxide ($C{O}_2$)

• The chemical formula is $C{O}_2$. This means one molecule of carbon dioxide contains 1 carbon atom and 2 oxygen atoms.
• Atomic mass of Carbon (C) = 12 u
• Atomic mass of Oxygen (O) = 16 u

Molecular mass of $C{O}_2$ = (Atomic mass of C × Number of C atoms) + (Atomic mass of O × Number of O atoms) Molecular mass of $C{O}_2$ = ($12\text{ u}\times 1$) + ($16\text{ u}\times 2$) Molecular mass of $C{O}_2$ = $12\text{ u}+32\text{ u}$ Molecular mass of $C{O}_2$ = 44 u

Why is Molecular Mass Important? Molecular mass is a fundamental property of covalent compounds. It is used in:

• Stoichiometry: Calculating the amounts of reactants and products in chemical reactions.
• Molar Mass: The molecular mass in grams (g) is numerically equal to the molar mass (grams per mole, g/mol) of the substance. This is crucial for understanding the quantity of matter in moles.
• Determining Physical Properties: Molecular mass influences properties like boiling point and melting point.

Remember

Molecular mass is always expressed in atomic mass units (u). It is the mass of a single molecule.

Exam-smart

You might be given the atomic masses of elements and asked to calculate the molecular mass of a compound like methane ($C{H}_4$), nitric acid ($HN{O}_3$), or sulfuric acid ($H_{2}S{O}_4$). Pay close attention to the chemical formula to ensure you count the correct number of atoms for each element.

Formula Unit Mass of Ionic Compounds

In ionic compounds, atoms do not exist as discrete, independent molecules. Instead, they form a rigid, three-dimensional structure called a crystal lattice, where positively charged ions (cations) and negatively charged ions (anions) are held together by strong electrostatic forces. The chemical formula of an ionic compound, such as $NaCl$ or $MgO$, represents the simplest whole-number ratio of these ions within the crystal lattice. This fundamental unit, representing the smallest whole-number ratio of ions, is called a formula unit.

The formula unit mass is the sum of the atomic masses of all the atoms present in one formula unit of an ionic compound. It is calculated in a similar way to molecular mass, by adding the atomic masses of the constituent elements, taking into account the number of atoms of each element as indicated by its chemical formula.

Example 9.6: Formula unit mass of sodium oxide ($N{a}_{2}O$)

• Atomic mass of Sodium ($Na$) = $23\text{ u}$
• Atomic mass of Oxygen ($O$) = $16\text{ u}$

The formula unit of sodium oxide is $N{a}_{2}O$, which means it contains 2 sodium atoms and 1 oxygen atom in its simplest ratio.

Formula unit mass of $N{a}_{2}O=(\text{Atomic mass of }Na\times 2)+(\text{Atomic mass of }O\times 1)$

Example 9.7: Formula unit mass of calcium nitrate ($Ca(N{O}_3{)}_2$)

• Atomic mass of Calcium ($Ca$) = $40\text{ u}$
• Atomic mass of Nitrogen ($N$) = $14\text{ u}$
• Atomic mass of Oxygen ($O$) = $16\text{ u}$

The formula unit of calcium nitrate is $Ca(N{O}_3{)}_2$. This indicates that in the simplest ratio, there is 1 calcium atom, and the nitrate group ($N{O}_3$) appears twice. Inside the nitrate group, there is 1 nitrogen atom and 3 oxygen atoms. Therefore, in one formula unit of $Ca(N{O}_3{)}_2$, there is:

• 1 Calcium atom
• $1\times 2=2$ Nitrogen atoms
• $3\times 2=6$ Oxygen atoms

Formula unit mass of $Ca(N{O}_3{)}_2=(\text{Atomic mass of }Ca\times 1)+\{(\text{Atomic mass of }N\times 1)+(\text{Atomic mass of }O\times 3)\}\times 2$

Calculation of Formula Unit Mass

To calculate the formula unit mass:

1. Identify the elements present in the ionic compound.
2. Determine the number of atoms of each element in one formula unit, as given by the chemical formula.
3. Find the atomic mass of each element from the periodic table.
4. Multiply the atomic mass of each element by the number of atoms of that element in the formula unit.
5. Sum up these values to get the formula unit mass.

Key Difference: Molecular Mass vs. Formula Unit Mass

Remember

Molecular mass applies to substances that exist as individual molecules, while formula unit mass applies to ionic compounds that form extended crystal structures rather than discrete molecules.

Exam-smart

In case/source-based questions, you might be given the chemical formula of an ionic compound and asked to calculate its formula unit mass. You will need to identify the elements, their counts in the formula unit, and their respective atomic masses to perform the calculation. For example, if given $Mg(OH{)}_2$, you would need to account for 1 Mg, 2 O, and 2 H atoms.