AP Chemistry Flashcards: Stoichiometry

What this deck covers

This deck focuses on Stoichiometry, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the molar mass of $H_2SO_4$?

Answer: 98.1 g/mol. 2H (2) + S (32) + 4O (64) = 98 g/mol total.

Flashcard 2: Calculate the percent composition of oxygen in $H_2O$.

Answer: 88.81%. Oxygen mass ($16$) divided by $H_2O$ mass ($18$) $\times 100\%$.

Flashcard 3: What is the concentration of a solution with 5 moles of solute in 2 L of solution?

Answer: 2.5 M. $\frac{5 \text{ moles}}{2 \text{ L}} = 2.5$ M concentration.

Flashcard 4: Find the moles of $H_2O$ in 36 grams.

Answer: 2 moles. $36 \div 18 = 2$ using $H_2O$ molar mass of 18 g/mol.

Flashcard 5: What is the formula to calculate the number of particles from moles?

Answer: $\text{particles} = \text{moles} \times 6.022 \times 10^{23}$. Multiplies moles by Avogadro's number for particle count.

Flashcard 6: Determine the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen.

Answer: $CH_2O$. Convert percentages to moles, then find simplest ratio.

Flashcard 7: Find the molar mass of $CO_2$.

Answer: 44 g/mol. Carbon (12) + 2 oxygen atoms (16 each) = 44 g/mol.

Flashcard 8: What is the formula for molarity?

Answer: $\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}$. Relates amount of solute to solution volume.

Flashcard 9: How do you calculate mass from moles?

Answer: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$. Multiplies moles by molar mass to get mass in grams.

Flashcard 10: State Avogadro's number.

Answer: $6.022 \times 10^{23}$. The number of particles in one mole of any substance.

Flashcard 11: What is the first step in a stoichiometry problem?

Answer: Write a balanced chemical equation. Ensures atom conservation for stoichiometric calculations.

Flashcard 12: How is theoretical yield calculated?

Answer: Use stoichiometry from the balanced equation. Assumes complete reaction with limiting reactant consumed.

Flashcard 13: What is the purpose of a mole ratio in stoichiometry?

Answer: To convert between moles of different substances. Uses coefficients from balanced equations for conversions.

Flashcard 14: Calculate the number of molecules in 2 moles of $H_2$.

Answer: $1.204 \times 10^{24}$ molecules. $2 \times 6.022 \times 10^{23}$ molecules per mole.

Flashcard 15: What is the stoichiometric coefficient for $H_2$ in $2H_2 + O_2 \rightarrow 2H_2O$?

Answer:

1. Coefficient shows 2 molecules react in balanced equation.

Flashcard 16: Which unit is used to express concentration in stoichiometry?

Answer: Molarity (M). Standard concentration unit for solution calculations.

Flashcard 17: How many moles of $H_2$ are needed to produce 2 moles of $NH_3$ in $N_2 + 3H_2 \rightarrow 2NH_3$?

Answer: 3 moles. 3:2 mole ratio from balanced equation coefficients.

Flashcard 18: What is the molecular formula for a compound with empirical formula $CH_2O$ and molar mass 180 g/mol?

Answer: $C_6H_{12}O_6$. $180 \div 30 = 6$, so multiply $CH_2O$ by 6.

Flashcard 19: What is the formula to calculate the number of moles from mass?

Answer: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$. Divides mass by molar mass to find amount in moles.

Flashcard 20: Determine the limiting reactant in $2H_2 + O_2 \rightarrow 2H_2O$ given 5 moles $H_2$ and 2 moles $O_2$.

Answer: $H_2$. Needs 2.5 moles $O_2$, but only 2 available, so $H_2$ limits.

Flashcard 21: What is the molar mass of $H_2SO_4$?

Answer: 98.1 g/mol. 2H (2) + S (32) + 4O (64) = 98 g/mol total.

Flashcard 22: What is the formula for percent yield?

Answer: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$. Compares actual to theoretical yield as a percentage.

Flashcard 23: Convert 3 moles of $NaCl$ to grams.

Answer: 175.5 grams. $3 \times 58.5 = 175.5$ using $NaCl$ molar mass.

Flashcard 24: Calculate the volume at STP for 0.5 moles of gas.

Answer: 11.2 L. $0.5 \times 22.4 = 11.2$ liters at STP.

Flashcard 25: What is empirical formula?

Answer: The simplest whole-number ratio of atoms. Shows simplest ratio without molecular complexity.

Flashcard 26: Identify the limiting reactant in 4 moles $N_2$ and 10 moles $H_2$ for $N_2 + 3H_2 \rightarrow 2NH_3$.

Answer: $H_2$. Need 12 moles $H_2$ but only have 10; $H_2$ limits.

Flashcard 27: Identify the unit for molar mass.

Answer: grams per mole (g/mol). Standard unit expressing mass per mole of substance.

Flashcard 28: Find the moles of $H_2O$ in 36 grams.

Answer: 2 moles. $36 \div 18 = 2$ using $H_2O$ molar mass of 18 g/mol.

Flashcard 29: What is the formula to calculate the number of particles from moles?

Answer: $\text{particles} = \text{moles} \times 6.022 \times 10^{23}$. Multiplies moles by Avogadro's number for particle count.

Flashcard 30: Determine the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen.

Answer: $CH_2O$. Convert percentages to moles, then find simplest ratio.