Why is the mole central to Virginia SOL Chemistry?

A mole is $6.02 \times 10^{23}$ representative particles (Avogadro's number). It bridges the laboratory scale (grams you weigh) and the particle scale (atoms, molecules or formula units). Almost every quantitative item runs through the mole, from percent composition to stoichiometry to molarity, so converting confidently between mass, moles and particles is the master skill of the course.

How do I find the molar mass and percent composition?

Add the atomic masses of every atom in the formula, taken from the periodic table, multiplying each by its subscript: $\text{CaCO}_3$ is $40.1 + 12.0 + 3(16.0) = 100.1$ g/mol. Percent composition is the mass of an element divided by the molar mass, times 100. In water, oxygen is $88.9\%$ by mass even though it is one of three atoms, because percent composition is by mass, not by atom count.

What can I change when balancing an equation, and why?

Only coefficients, never subscripts. A coefficient multiplies the whole formula and changes how many units there are, which conserves the substance; changing a subscript would create a different compound. Balance one element at a time, leave hydrogen and oxygen until last (oxygen very last in combustion), and reduce to the smallest whole numbers. The balanced coefficients are the mole ratios for stoichiometry.

How do I do a mass-to-mass stoichiometry calculation?

Run a three-step chain: convert the given mass to moles (divide by molar mass), apply the mole ratio from the balanced coefficients, then convert the target moles to mass (multiply by molar mass). For a gas at STP, one mole occupies 22.4 L, so you can convert moles of gas to volume. A pure mole-to-mole question just applies the ratio.

What are limiting reactants and percent yield?

The limiting reactant runs out first and sets the maximum product; the excess reactant is left over. Find the limiting reactant by converting each reactant to moles and seeing which makes the least product against the mole ratio. The theoretical yield is the maximum from the limiting reactant, and percent yield is the actual yield divided by the theoretical yield times 100. Percent yield cannot exceed 100 percent.

VirginiaChemistry

Virginia SOL Chemistry molar relationships and chemical reactions: a complete skills guide to the mole, percent composition, balancing equations, reaction types and stoichiometry

A deep-dive Virginia SOL Chemistry guide to molar relationships and chemical reactions (CH.3): the mole and molar mass, percent composition and empirical formulas, balancing equations and conservation of mass, the five reaction types, mole-ratio stoichiometry with gas volumes, and limiting reactants and percent yield, with the periodic table and SOL exam technique.

Generated by Claude Opus 4.817 min readCH.3Updated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

Why this is the quantitative spine of the course
The mole and molar mass
Composition and formulas
Balancing and reaction types
Stoichiometry and yield
Check your knowledge

Why this is the quantitative spine of the course

The third reporting category, Molar Relationships, Nomenclature, Chemical Equations and Reactions, is where chemistry becomes quantitative. The mole connects mass to particle counts, balanced equations give mole ratios, and stoichiometry turns those ratios into masses and volumes you can measure. These skills recur in solutions, neutralization and energy calculations later in the course. This guide ties together the matching dot-point pages, each with its own practice: the mole and molar mass, percent composition and empirical formulas, balancing equations and conservation of mass, types of chemical reactions, stoichiometry and the mole ratio, and limiting reactants and percent yield.

The mole and molar mass

A mole is $6.02 \times 10^{23}$ particles. The molar mass is the sum of the atomic masses in a formula, in g/mol. The links are:

\text{moles} = \frac{\text{mass}}{\text{molar mass}}, \qquad \text{particles} = \text{moles} \times (6.02 \times 10^{23})

Treat mass, moles and particles as a chain you can travel either way. The commonest slip is confusing molecules with atoms: one mole of $\text{O}_2$ has $6.02 \times 10^{23}$ molecules but twice as many atoms.

Composition and formulas

Percent composition is the mass of an element over the molar mass, times $100$ . Working backward, the empirical formula is the simplest whole-number ratio: assume $100$ g, convert each element to moles, and divide by the smallest. The molecular formula is a whole-number multiple of the empirical formula, found by dividing the actual molar mass by the empirical-formula mass.

Balancing and reaction types

Balance equations by adjusting coefficients only so atoms are conserved (the law of conservation of mass), leaving hydrogen and oxygen until last. The five reaction types are synthesis ( $\text{A} + \text{B} \rightarrow \text{AB}$ ), decomposition ( $\text{AB} \rightarrow \text{A} + \text{B}$ ), single replacement ( $\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$ ), double replacement ( $\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$ ), and combustion (fuel plus oxygen gives carbon dioxide and water). Use an activity series for single replacement and the precipitate, gas, or water rule for double replacement.

Stoichiometry and yield

The balanced coefficients are the mole ratio. For mass-to-mass problems, run the chain: grams to moles, mole ratio, moles to grams. At STP, one mole of gas occupies $22.4$ L. When one reactant runs out first, it is the limiting reactant and sets the theoretical yield; the percent yield is the actual yield over the theoretical yield, times $100$ .

Step	Operation
1	Convert given mass to moles: divide by molar mass
2	Apply the mole ratio from the balanced coefficients
3	Convert target moles to mass: multiply by molar mass

A full mass-to-mass problem

Given $2\,\text{Mg} + \text{O}_2 \rightarrow 2\,\text{MgO}$ , calculate the mass of magnesium oxide formed when $4.86$ g of magnesium burns completely. (Molar masses: Mg $24.3$ , MgO $40.3$ g/mol.)

step 1 Mass of magnesium to moles

$\text{moles Mg} = \dfrac{4.86\ \text{g}}{24.3\ \text{g/mol}} = 0.200$ mol.

step 2 Apply the mole ratio

The ratio $\text{Mg} : \text{MgO}$ is $2 : 2$ , so $0.200$ mol Mg gives $0.200$ mol MgO.

step 3 Moles of magnesium oxide to mass

$\text{mass MgO} = 0.200\ \text{mol} \times 40.3\ \text{g/mol} = 8.06$ g.

step 4 Sense-check

The product gains the mass of the oxygen that combined, so the magnesium oxide mass exceeds the magnesium mass, which it does.

The mole ( $6.02 \times 10^{23}$ particles) links mass and particle counts through the molar mass: moles equal mass over molar mass, and particles equal moles times Avogadro's number. Percent composition is mass of element over molar mass times $100$ ; the empirical formula is the simplest mole ratio and the molecular formula a whole-number multiple of it. Balance equations by adjusting coefficients only so mass is conserved; the coefficients are mole ratios. The five reaction types are synthesis, decomposition, single replacement, double replacement and combustion. For stoichiometry, run grams to moles, mole ratio, moles to grams, and use $22.4$ L per mole for a gas at STP. The limiting reactant sets the theoretical yield, and percent yield is actual over theoretical times $100$ , capped at $100\%$ .

Check your knowledge

Attempt these under timed conditions, then check the solutions.

Calculate the molar mass of $\text{Ca(OH)}_2$ . (2 marks)
Calculate the percent by mass of carbon in $\text{CO}_2$ (molar mass $44.0$ ). (2 marks)
Balance: $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$ . (2 marks)
Classify the reaction $\text{Zn} + 2\,\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$ . (1 mark)
For $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$ , how many moles of ammonia form from $9.0$ mol of hydrogen? (2 marks)
A reaction has a theoretical yield of $50.0$ g and an actual yield of $42.0$ g. Calculate the percent yield. (2 marks)

Solutions

Step 1: Q1. Calculate the molar mass of Ca(OH)2

Add the atomic masses of every atom in the formula, multiplying each by its subscript. The subscript $2$ outside the bracket applies to the whole $\text{OH}$ group:

M(\text{Ca(OH)}_2) = 40.1 + 2(16.0 + 1.0) = 40.1 + 2(17.0) = 74.1 \text{ g/mol}

Step 2: Q2. Calculate percent by mass of carbon in CO2

Percent composition is the mass of the element divided by the molar mass of the compound, times $100$ . Carbon has one atom at $12.0$ g/mol; the molar mass of $\text{CO}_2$ is $44.0$ g/mol:

\frac{12.0}{44.0} \times 100 = 27.3\%

Step 3: Q3. Balance the iron oxide equation

Count the atoms on each side and adjust coefficients. Placing $2$ in front of $\text{Fe}_2\text{O}_3$ gives $4$ Fe and $6$ O on the right, so balance the left with $4$ Fe and $3\,\text{O}_2$ :

4\,\text{Fe} + 3\,\text{O}_2 \rightarrow 2\,\text{Fe}_2\text{O}_3 \quad (4 \text{ Fe and } 6 \text{ O on each side})

Step 4: Q4. Classify the reaction type

Zinc is a free element that takes the place of hydrogen in $\text{HCl}$ . One element displacing another from a compound is the definition of single replacement:

\text{Single replacement}

Step 5: Q5. Apply the mole ratio for stoichiometry

From the balanced equation, $\text{H}_2 : \text{NH}_3 = 3 : 2$ . Multiply the given moles of hydrogen by the ratio $\frac{2}{3}$ :

9.0 \times \frac{2}{3} = 6.0 \text{ mol of ammonia}

Step 6: Q6. Calculate percent yield

Divide the actual yield by the theoretical yield and multiply by $100$ . Percent yield cannot exceed $100\%$ :

\frac{42.0}{50.0} \times 100 = 84.0\%

Sources & how we know this

2018 Science Standards of Learning - Chemistry — Virginia Department of Education (2018)
Chemistry Curriculum Framework — Virginia Department of Education (2018)