Moles and Molar Mass - AP Chemistry

A pure sample contains $3.00\ \mathrm{mol}$ of sodium atoms, $\mathrm{Na(s)}$ . Using Avogadro's number, $6.02\times 10^{23}\ \mathrm{particles\ mol^{-1}}$ , how many sodium atoms are in the sample?
Answer: $1.81\times 10^{24}\ \mathrm{atoms}$ This problem tests your ability to use moles and molar mass concepts, specifically Avogadro's number. Avogadro's number (6.02 × 10²³ particles/mol) connects moles to the actual number of particles (atoms, molecules, or formula units). To find the number of atoms from moles, multiply: atoms = moles × Avogadro's number. Here, 3.00 mol × 6.02 × 10²³ atoms/mol = 1.81 × 10²⁴ atoms. A common mistake is dividing instead of multiplying (3.00 ÷ 6.02 × 10²³), which would give an impossibly small number of atoms. The strategy is to ensure units cancel properly: start with moles, multiply by Avogadro's number (atoms/mol), and confirm moles cancel leaving atoms.
A sample contains $2.00\text{ mol}$ of pure ammonia, $\text{NH}_3$ . What is the mass of the sample? (Molar mass of $\text{NH}_3=17.0\text{ g/mol}$ .)
Answer: $34.0\text{ g}$ This problem requires applying moles and molar mass relationships. Molar mass tells us the mass of one mole of a substance, so to find the mass of multiple moles, we multiply the number of moles by the molar mass. The calculation follows: mass = moles × molar mass. For ammonia: mass = 2.00 mol × 17.0 g/mol = 34.0 g. A tempting mistake would be dividing molar mass by moles (17.0 ÷ 2.00 = 8.50 g, choice D), which incorrectly treats molar mass as a total rather than a per-mole value. Always multiply moles by molar mass when finding mass—you're scaling up from the per-mole mass to the total mass.
A $40.0\ \mathrm{g}$ sample of pure sodium hydroxide, $\mathrm{NaOH}$ , is measured. Given that the molar mass of $\mathrm{NaOH}$ is $40.0\ \mathrm{g\ mol^{-1}}$ , how many moles of $\mathrm{NaOH}$ does the sample contain?
Answer: $1.00\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass links mass to moles by dividing grams by grams per mole for NaOH with 40.0 g/mol. This molar mass is summed from sodium, oxygen, and hydrogen atoms. Avogadro's number connects moles to particles, though mass-to-moles is key here. A common distractor is choice B, 40.0 mol, from multiplying 40.0 by 40.0 instead of dividing, inverting the conversion. Divide for moles from mass. A transferable strategy is to write the unit you want (moles) and use molar mass as the bridge by dividing grams by grams per mole.
A sample contains $1.20\times10^{24}$ molecules of pure ammonia, $\text{NH}_3$ . How many moles of $\text{NH}_3$ are present? (Use $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$ .)
Answer: $1.99\ \text{mol}$ This question tests the skill of moles and molar mass. Avogadro's number (6.02×10²³ particles/mol) converts between particles and moles using: moles = particles ÷ Avogadro's number. With 1.20×10²⁴ molecules of NH₃, we divide: 1.20×10²⁴ ÷ 6.02×10²³ mol⁻¹ = 1.99 mol (keeping significant figures). The molar mass of NH₃ would only be needed if converting to mass, but here we're converting molecules to moles. A common mistake (choice C, 2.00 mol) results from improper rounding—while 1.20/6.02 ≈ 0.199×10 = 1.99, students might round to 2.00 too early. To convert particles to moles accurately, divide by Avogadro's number and maintain proper significant figures throughout the calculation.
A sample contains $0.750\text{ mol}$ of pure iron(III) oxide, $\text{Fe}_2\text{O}_3$ . What is the mass of the sample? (Molar mass of $\text{Fe}_2\text{O}_3=160.0\text{ g/mol}$ .)
Answer: $120\text{ g}$ This problem requires applying moles and molar mass relationships. Molar mass tells us that one mole of Fe₂O₃ has a mass of 160.0 grams, so we can scale this up for any number of moles. To find mass from moles, we multiply: mass = moles × molar mass. For iron(III) oxide: mass = 0.750 mol × 160.0 g/mol = 120 g. A tempting error would be adding the values (0.750 + 160.0 = 160.75 ≈ 160, choice C), which incorrectly treats molar mass as something to add rather than multiply. Remember to multiply moles by molar mass—you're finding how many grams correspond to your specific number of moles.
A student has $0.300\ \mathrm{mol}$ of pure glucose, $\mathrm{C_6H_{12}O_6}$ . (Atomic masses: $\mathrm{C}=12.0$ , $\mathrm{H}=1.0$ , $\mathrm{O}=16.0$ .) What is the mass of the glucose sample?
Answer: $54.0\ \mathrm{g}$ This question tests the skill of moles and molar mass. Molar mass links mass to amount by mass = moles × molar mass, with glucose at 6 × 12.0 + 12 × 1.0 + 6 × 16.0 = 180 g/mol. This bridges the given moles to mass. Avogadro's number, 6.02 × 10²³, links moles to particles, essential for particle-level understanding. A tempting distractor is choice A, 18.0 g, from a common unit-conversion error of using water's molar mass instead of glucose's. A transferable strategy is to write the unit you want and use molar mass as the bridge.
A sample contains $1.204\times10^{24}$ formula units of pure sodium fluoride, $\text{NaF}$ . What amount of $\text{NaF}$ (in moles) is present? (Use $N_A=6.02\times10^{23}\ \text{mol}^{-1}$ .)
Answer: $2.00\ \text{mol}$ This problem tests moles and molar mass concepts by converting formula units to moles using Avogadro's number. One mole contains 6.02×10²³ formula units (the particle count for ionic compounds). To find moles: 1.204×10²⁴ formula units ÷ 6.02×10²³ formula units/mol = 2.00 mol. A common mistake would be to use 1.204×10²³ instead of 1.204×10²⁴, which would give 0.200 mol. The key strategy is to recognize that "formula units" is used for ionic compounds and works exactly like molecules in calculations with Avogadro's number.
A student has a pure sample of sodium chloride, $\mathrm{NaCl(s)}$ , with a mass of $58.5\ \mathrm{g}$ . The molar mass of $\mathrm{NaCl}$ is $58.5\ \mathrm{g\ mol^{-1}}$ . How many formula units of $\mathrm{NaCl}$ are in the sample? (Use $6.02\times 10^{23}\ \mathrm{particles\ mol^{-1}}$ .)
Answer: $6.02\times 10^{23}\ \mathrm{formula\ units}$ This problem combines moles and molar mass concepts to find the number of formula units. First, we convert mass to moles: 58.5 g ÷ 58.5 g/mol = 1.00 mol. Then, we use Avogadro's number to convert moles to formula units: 1.00 mol × 6.02 × 10²³ formula units/mol = 6.02 × 10²³ formula units. A student might incorrectly multiply the mass by Avogadro's number directly (58.5 × 6.02 × 10²³), forgetting to first convert to moles. The systematic approach is: mass → moles (using molar mass) → particles (using Avogadro's number).
A student measures out $44.0\ \text{g}$ of pure carbon dioxide, $\text{CO}_2$ . (Molar mass of $\text{CO}_2=44.0\ \text{g/mol}$ .) How many moles of $\text{CO}_2$ are present?
Answer: $1.00\ \text{mol}$ This question tests the skill of moles and molar mass. Molar mass provides the conversion factor between mass and moles through the relationship: moles = mass ÷ molar mass. With 44.0 g of CO₂ and a molar mass of 44.0 g/mol, we calculate: 44.0 g ÷ 44.0 g/mol = 1.00 mol. Avogadro's number would be necessary to find the number of molecules, but here we only need moles. A tempting wrong answer (choice A, 0.0227 mol) results from incorrectly inverting the calculation and dividing 1 by 44.0 instead of dividing 44.0 by 44.0. To solve mole conversions accurately, set up the calculation so that grams cancel out, leaving moles as your final unit.
A student has $0.250\ \mathrm{mol}$ of pure calcium carbonate, $\mathrm{CaCO_3}$ . Given that the molar mass of $\mathrm{CaCO_3}$ is $100.0\ \mathrm{g\ mol^{-1}}$ , what is the mass of the sample?
Answer: $25.0\ \mathrm{g}$ This question tests the skill of moles and molar mass. Molar mass links the mass of a substance to the amount in moles, enabling us to find the mass by multiplying the number of moles by the molar mass in grams per mole. For ionic compounds like CaCO3, the molar mass is calculated from the periodic table values of its atoms, facilitating conversions between lab-scale masses and stoichiometric amounts. Avogadro's number connects moles to the number of formula units or molecules, providing a way to quantify particles from moles, although this problem emphasizes moles-to-mass conversion. A common distractor is choice A, 2.50 g, which arises from mistakenly dividing 0.250 by 100.0 instead of multiplying, a frequent unit-conversion error when confusing the formula. Remember to verify whether you need to multiply or divide based on the desired unit. A transferable strategy is to write the unit you want (grams) and use molar mass as the bridge by multiplying moles by grams per mole.
A $9.00\ \mathrm{g}$ sample of pure aluminum, $\mathrm{Al}$ , is obtained. (Atomic mass: $\mathrm{Al}=27.0$ .) How many moles of aluminum atoms are present?
Answer: $0.333\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass links mass to amount via moles = mass / molar mass, with Al at 27.0 g/mol giving 0.333 mol for 9.00 g. This simple calculation uses atomic mass directly. Avogadro's number, 6.02 × 10²³, links moles to particles for atom counts. A tempting distractor is choice B, 0.250 mol, from a common unit-conversion error of using 36 g/mol instead of 27. A transferable strategy is to write the unit you want and use molar mass as the bridge.
A student has $0.500\ \mathrm{mol}$ of pure ammonia, $\mathrm{NH_3(s)}$ . The molar mass of $\mathrm{NH_3}$ is $17.0\ \mathrm{g\ mol^{-1}}$ . What is the mass of the ammonia sample?
Answer: $8.50\ \mathrm{g}$ This problem requires using moles and molar mass to convert from moles to mass. The molar mass of NH₃ is 17.0 g/mol, which tells us the mass of one mole. To find the mass of 0.500 mol, we multiply: 0.500 mol × 17.0 g/mol = 8.50 g. A student might mistakenly divide the molar mass by moles (17.0 ÷ 0.500 = 34.0), which gives twice the molar mass rather than half. The reliable method is dimensional analysis: start with moles, multiply by molar mass (g/mol), and verify that mol cancels to leave grams.
A sample of pure magnesium chloride, $\mathrm{MgCl_2}$ , has a mass of $9.50\ \mathrm{g}$ . Given that the molar mass of $\mathrm{MgCl_2}$ is $95.0\ \mathrm{g\ mol^{-1}}$ , how many moles of $\mathrm{MgCl_2}$ are in the sample?
Answer: $0.100\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass serves as the conversion factor that links the mass of a substance in grams to the amount in moles, allowing us to calculate moles from mass by dividing the given mass by the molar mass. For compounds like MgCl2, the molar mass is the sum of the atomic masses of its constituent elements, providing a direct bridge between macroscopic measurements and microscopic quantities. Avogadro's number, 6.022 × $10^{23}$ particles per mole, further connects the amount in moles to the number of individual particles, though this question focuses on mass-to-moles conversion. A tempting distractor might be choice A, 0.0100 mol, which results from incorrectly dividing 9.50 by 950 instead of 95.0, a common error in handling decimal places during unit conversion. Always double-check the placement of decimals when performing divisions with molar mass. To solve these problems reliably, write the unit you want (moles) and use molar mass as the bridge by dividing mass by grams per mole.
A pure sample contains $0.250\ \mathrm{mol}$ of sodium carbonate, $\mathrm{Na_2CO_3}$ . (Atomic masses: $\mathrm{Na}=23.0$ , $\mathrm{C}=12.0$ , $\mathrm{O}=16.0$ .) What is the mass of the sample?
Answer: $26.5\ \mathrm{g}$ This question tests the skill of moles and molar mass. Molar mass links mass to amount by providing the grams per mole, so mass = moles × molar mass for a given substance. For Na2CO3, it's calculated as 2 × 23.0 + 12.0 + 3 × 16.0 = 106 g/mol. Avogadro's number, 6.02 × 10²³, connects moles to particles, allowing conversion between the amount in moles and the actual number of formula units or molecules. A tempting distractor is choice C, 53.0 g, from a common unit-conversion error of halving the molar mass incorrectly, perhaps miscounting the sodium atoms. A transferable strategy is to write the unit you want and use molar mass as the bridge.
A sample contains $2.50\ \mathrm{mol}$ of pure sulfur atoms, $\mathrm{S(s)}$ . Using Avogadro's number, $6.02\times 10^{23}\ \mathrm{particles\ mol^{-1}}$ , how many sulfur atoms are in the sample?
Answer: $1.51\times 10^{24}\ \mathrm{atoms}$ This problem tests your understanding of moles and molar mass, specifically using Avogadro's number. To convert from moles to atoms, we multiply by Avogadro's number: 2.50 mol × 6.02 × 10²³ atoms/mol = 1.51 × 10²⁴ atoms. The calculation involves multiplying 2.50 × 6.02 = 15.05, then adjusting for scientific notation to get 1.505 × 10²⁴ (rounded to 1.51 × 10²⁴). A student might divide instead (2.50 ÷ 6.02 × 10²³), getting an impossibly small number. The strategy is to remember that Avogadro's number is a conversion factor: multiply by it to go from moles to particles.
A $23.0\ \mathrm{g}$ sample of pure ethanol, $\mathrm{C_2H_6O}$ , is used. (Atomic masses: $\mathrm{C}=12.0$ , $\mathrm{H}=1.0$ , $\mathrm{O}=16.0$ .) How many moles of ethanol are present?
Answer: $0.500\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass links mass to amount as moles = mass / molar mass, with ethanol at 46.0 g/mol giving 0.500 mol. This conversion is fundamental. Avogadro's number, 6.02 × 10²³, links moles to particles. A tempting distractor is choice A, 0.250 mol, from a common unit-conversion error of halving the mass. A transferable strategy is to write the unit you want and use molar mass as the bridge.
A sample of pure sulfur, $\mathrm{S}$ , has a mass of $6.40\ \mathrm{g}$ . Given that the molar mass of sulfur is $32.0\ \mathrm{g\ mol^{-1}}$ , how many moles of $\mathrm{S}$ are present?
Answer: $0.200\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass connects mass in grams to moles by dividing for elemental sulfur at 32.0 g/mol. This value is the atomic mass of sulfur. Avogadro's number links moles to atoms, but this focuses on mass-to-moles. A tempting distractor is choice E, 6.40 mol, from multiplying 6.40 by 32.0 instead of dividing, a unit-conversion mix-up. Always divide mass by molar mass. A transferable strategy is to write the unit you want (moles) and use molar mass as the bridge by dividing grams by grams per mole.
A sample contains $6.022\times 10^{22}$ molecules of pure methane, $\mathrm{CH_4}$ . How many moles of $\mathrm{CH_4}$ are in the sample? (Avogadro's number: $6.022\times 10^{23}\ \mathrm{mol^{-1}}$ .)
Answer: $0.100\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass relates mass to moles, but here we convert molecules to moles using Avogadro's number. Avogadro's number bridges by dividing particles by 6.022 × $10^{23}$ per mole for CH4. This finds moles from molecule count. A common distractor is choice E, 6.022 × $10^{22}$ mol, from not dividing by Avogadro's number, treating the number as moles directly. Divide to get moles. A transferable strategy is to write the unit you want (moles) and use Avogadro's number as the bridge by dividing particles by particles per mole.
A student has $0.250\ \text{mol}$ of pure sodium chloride, $\text{NaCl}$ . (Molar mass of $\text{NaCl}=58.5\ \text{g/mol}$ .) What is the mass of the sample?
Answer: $14.6\ \text{g}$ This question tests the skill of moles and molar mass. Molar mass serves as the conversion factor linking moles to mass, with the relationship: mass = moles × molar mass. Given 0.250 mol of NaCl with molar mass 58.5 g/mol, we multiply: 0.250 mol × 58.5 g/mol = 14.6 g. Avogadro's number connects moles to particles but isn't needed here since we're converting between moles and grams. A tempting incorrect answer (choice C, 58.5 g) comes from confusing the molar mass itself with the actual mass of the sample—remember that 58.5 g is the mass of 1 mole, not 0.250 mol. When converting from moles to grams, always multiply moles by molar mass to ensure proper unit cancellation.
A $18.0\ \mathrm{g}$ sample of pure water, $\mathrm{H_2O}$ , is analyzed. Given that the molar mass of $\mathrm{H_2O}$ is $18.0\ \mathrm{g\ mol^{-1}}$ , how many moles of $\mathrm{H_2O}$ are present?
Answer: $1.00\ \mathrm{mol}$ This question tests the skill of moles and molar mass. Molar mass links the mass in grams to the amount in moles, calculated by dividing mass by molar mass for substances like H2O. The molar mass of water is 18.0 g/mol, derived from hydrogen and oxygen atomic masses, enabling precise conversions. Avogadro's number connects moles to the number of molecules, though this question focuses on mass-to-moles. A tempting distractor is choice C, 18.0 mol, from incorrectly multiplying 18.0 by 18.0 instead of dividing, a unit-conversion error when mixing up operations. Always divide mass by molar mass for moles. A transferable strategy is to write the unit you want (moles) and use molar mass as the bridge by dividing grams by grams per mole.

Start this quiz