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 first step in a stoichiometry problem?

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

Flashcard 2: 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 3: Calculate the percent composition of oxygen in $H_2O$.

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

Flashcard 4: Find the molar mass of $CO_2$.

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

Flashcard 5: Identify the unit for molar mass.

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

Flashcard 6: Which unit is used to express concentration in stoichiometry?

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

Flashcard 7: Calculate the molarity of a solution with 10 moles in 5 L.

Answer: 2 M. $\frac{10 \text{ moles}}{5 \text{ L}} = 2$ M concentration.

Flashcard 8: 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 9: 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 10: 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 11: Which unit is used to express concentration in stoichiometry?

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

Flashcard 12: 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 13: 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 14: 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 15: 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 16: How is actual yield different from theoretical yield?

Answer: Actual yield is the measured amount of product obtained. Theoretical assumes perfect conditions; actual is experimental.

Flashcard 17: How is theoretical yield calculated?

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

Flashcard 18: 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 19: 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 20: 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 21: 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 22: Convert 3 moles of $NaCl$ to grams.

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

Flashcard 23: Calculate the percent composition of oxygen in $H_2O$.

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

Flashcard 24: 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 25: 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 26: 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 27: Determine the moles of $NaOH$ needed to neutralize $1$ mole of $HCl$.

Answer: $1$ mole. $1:1$ mole ratio from balanced acid-base equation.

Flashcard 28: 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 29: Define molar mass.

Answer: The mass of one mole of a substance in grams. Equals the atomic/molecular weight expressed in g/mol.

Flashcard 30: Find the volume of gas at STP for 1 mole.

Answer: 22.4 L. Standard molar volume at 0°C and 1 atm pressure.

Flashcard 31: What is the first step in a stoichiometry problem?

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

Flashcard 32: 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 33: Calculate the mass of 0.5 moles of $C_6H_{12}O_6$.

Answer: 90 grams. $0.5 \times 180 = 90$ using glucose molar mass.

Flashcard 34: How is theoretical yield calculated?

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

Flashcard 35: What is the excess reactant in $2H_2 + O_2 \rightarrow 2H_2O$ with 5 moles $H_2$ and 2 moles $O_2$?

Answer: $O_2$. Remains after limiting reactant is completely consumed.

Flashcard 36: 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 37: Determine the moles of $NaOH$ needed to neutralize 1 mole of $HCl$.

Answer: 1 mole. 1:1 mole ratio from balanced acid-base equation.

Flashcard 38: Convert 0.25 moles of $Mg$ to atoms.

Answer: $1.505 \times 10^{23}$ atoms. $0.25 \times 6.022 \times 10^{23}$ atoms per mole.

Flashcard 39: 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 40: What is empirical formula?

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

Flashcard 41: What is the percent yield if the theoretical yield is 100g and actual yield is 80g?

Answer: 80%. $\frac{80}{100} \times 100\% = 80\%$ efficiency.

Flashcard 42: Which unit is used to express concentration in stoichiometry?

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

Flashcard 43: 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 44: 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 45: Convert 3 moles of $NaCl$ to grams.

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

Flashcard 46: 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 47: Find the molar mass of $CO_2$.

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

Flashcard 48: Calculate the mass of 0.5 moles of $C_6H_{12}O_6$.

Answer: 90 grams. $0.5 \times 180 = 90$ using glucose molar mass.

Flashcard 49: 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 50: 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 51: 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 52: Calculate the mass of 0.5 moles of $C_6H_{12}O_6$.

Answer: 90 grams. $0.5 \times 180 = 90$ using glucose molar mass.

Flashcard 53: How is actual yield different from theoretical yield?

Answer: Actual yield is the measured amount of product obtained. Theoretical assumes perfect conditions; actual is experimental.

Flashcard 54: 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 55: Identify the unit for molar mass.

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

Flashcard 56: Find the volume of gas at STP for 1 mole.

Answer: 22.4 L. Standard molar volume at 0°C and 1 atm pressure.

Flashcard 57: Identify the unit for molar mass.

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

Flashcard 58: How is theoretical yield calculated?

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

Flashcard 59: What is the excess reactant in $2H_2 + O_2 \rightarrow 2H_2O$ with 5 moles $H_2$ and 2 moles $O_2$?

Answer: $O_2$. Remains after limiting reactant is completely consumed.

Flashcard 60: What is the excess reactant in $2H_2 + O_2 \rightarrow 2H_2O$ with 5 moles $H_2$ and 2 moles $O_2$?

Answer: $O_2$. Remains after limiting reactant is completely consumed.

Flashcard 61: 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 62: Determine the moles of $NaOH$ needed to neutralize 1 mole of $HCl$.

Answer: 1 mole. $1:1$ mole ratio from balanced acid-base equation.

Flashcard 63: Determine the empirical formula for a compound with 50% sulfur and 50% oxygen by mass.

Answer: $SO_2$. Equal mass percentages give 1:2 atomic ratio S:O.

Flashcard 64: 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 65: 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 66: 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 67: 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 68: 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 69: 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 70: 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 71: How many grams of $O_2$ are needed to react with 4 moles of $C$ in $C + O_2 \rightarrow CO_2$?

Answer: 128 grams. $4 \times 32 = 128$ grams using 1:1 mole ratio.

Flashcard 72: 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 73: What is empirical formula?

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

Flashcard 74: What is the percent yield if the theoretical yield is 100g and actual yield is 80g?

Answer: 80%. $\frac{80}{100} \times 100\% = 80\%$ efficiency.

Flashcard 75: State Avogadro's number.

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

Flashcard 76: 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 77: State Avogadro's number.

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

Flashcard 78: 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 79: 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 80: What is the first step in a stoichiometry problem?

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

Flashcard 81: Determine the empirical formula for a compound with 50% sulfur and 50% oxygen by mass.

Answer: $SO_2$. Equal mass percentages give 1:2 atomic ratio S:O.

Flashcard 82: 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 83: Convert 3 moles of $NaCl$ to grams.

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

Flashcard 84: Find the molar mass of $CO_2$.

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

Flashcard 85: 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 86: 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 87: 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 88: 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 89: Find the volume of 3 moles of gas at STP.

Answer: 67.2 L. $3 \times 22.4 = 67.2$ liters at standard conditions.

Flashcard 90: 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 91: How is actual yield different from theoretical yield?

Answer: Actual yield is the measured amount of product obtained. Theoretical assumes perfect conditions; actual is experimental.

Flashcard 92: Calculate the molarity of a solution with 10 moles in 5 L.

Answer: 2 M. $\frac{10 \text{ moles}}{5 \text{ L}} = 2$ M concentration.

Flashcard 93: 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 94: Define molar mass.

Answer: The mass of one mole of a substance in grams. Equals the atomic/molecular weight expressed in g/mol.

Flashcard 95: 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 96: Determine the empirical formula for a compound with 50% sulfur and 50% oxygen by mass.

Answer: $SO_2$. Equal mass percentages give 1:2 atomic ratio S:O.

Flashcard 97: What is the percent yield if the theoretical yield is 100g and actual yield is 80g?

Answer: 80%. $\frac{80}{100} \times 100\% = 80\%$ efficiency.

Flashcard 98: 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 99: What is empirical formula?

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

Flashcard 100: 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.