Mole Quantities
You have 2.00 g of H₂, 2.00 g of N₂, and 2.00 g of O₂. Which sample contains the greatest number of molecules? Explain your reasoning in one sentence.
Hydrogen-- smallest molar mass, so it takes the most moles of Hydrogen to get to 2g.
A pure sample of sodium sulfate, Na₂SO₄, is split into four portions:
• Sample A: 2.00 g of Na₂SO₄ crystals
• Sample B: 2.00 g of Na₂SO₄, finely ground into a powder
• Sample C: 5.00 g of Na₂SO₄ crystals
• Sample D: 5.00 g of Na₂SO₄ that has been melted and allowed to cool back into a solid (no decomposition)
Rank the four samples from greatest to smallest percent by mass of sodium. If any are equal, state that clearly and justify your answer in words (no calculations).
All four samples have exactly the same percent by mass of sodium.
Reason
Percent composition is a property of the chemical formula itself. Changing the sample size, grinding it, or melting and re-solidifying it (without decomposition) does not change the ratio of Na, S, and O in Na₂SO₄.
A compound is found to be 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen by mass. A student sets up the mole ratio and gets C:3.33, H:6.67, O:3.47. The student immediately writes the empirical formula as C₃H₇O₄.
Explain the student’s mistake and state the correct empirical formula.
CH₂O
A 1.50 g sample of a hydrate of calcium chloride is heated until only 0.90 g of anhydrous CaCl₂ remains. Determine the mass of water lost and the percent by mass of water in the hydrate.
Mass of water lost
1.50 g − 0.90 g = 0.60 g
Percent water
(0.60 g ÷ 1.50 g) × 100 = 40.0 percent water
A sample contains 4.50×10²² molecules of nitrogen gas, N₂. Convert this quantity to moles using correct scientific notation and appropriate significant figures.
n ≈ 7.47×10⁻² mol = .0747 mol
n = (4.50×10²² molecules) ÷ (6.022×10²³ molecules/mol)
n ≈ 7.47×10⁻² mol
With sig figs from the given value (3 sig figs):
n ≈ 7.47×10⁻² mol
You have 1.00 mol of each ionic compound: Na₂SO₄, CaCl₂, and Al(NO₃)₃. Which sample contains the greatest number of formula units? Explain your reasoning.
All three contain the same number of formula units.
Reason
A mole is defined as 6.022×10²³ formula units regardless of how many ions or polyatomic groups appear in each formula.
A compound contains only carbon, hydrogen, and oxygen. Two different students analyze two different pure samples of the same compound.
Student A’s sample is 1.00 g and contains 0.400 g of carbon.
Student B’s sample is 2.50 g and contains 1.00 g of carbon.
Both students claim they found the percent composition of carbon in the compound.
Are their results consistent with each other? Explain why or why not.
Yes, the two results are consistent.
Reason
Percent composition does not depend on how much of the compound you have.
A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. A student converts the percentages to moles and gets the ratio C:4.35, H:10.8, O:2.18. The student decides the empirical formula is C₄H₁₁O₂.
Explain why this is not correct and determine the proper empirical formula.
C₂H₅O.
The student rounded too early and did not correctly reduce the mole ratios to the simplest whole numbers.
From the student’s values:
C: 4.35, H: 10.8, O: 2.18
Divide all by the smallest, 2.18:
C ≈ 4.35 ÷ 2.18 ≈ 2
H ≈ 10.8 ÷ 2.18 ≈ 5
O ≈ 2.18 ÷ 2.18 = 1
So the simplest whole-number ratio is C₂H₅O.
A 2.50 g sample of a cobalt chloride hydrate is heated strongly until its mass stops changing. After heating, the mass is 1.60 g. The anhydrous salt is CoCl₂.
Determine the empirical formula of the hydrate, CoCl₂·xH₂O.
(Show enough reasoning to justify x.)
CoCl₂·4H₂O
Mass of water lost:
2.50 g − 1.60 g = 0.90 g H₂O
Moles of CoCl₂:
M(CoCl₂) ≈ 129.8 g/mol
n(CoCl₂) = 1.60 g ÷ 129.8 g/mol ≈ 0.0123 mol
Moles of H₂O:
M(H₂O) ≈ 18.0 g/mol
n(H₂O) = 0.90 g ÷ 18.0 g/mol ≈ 0.0500 mol
Mole ratio water to salt:
0.0500 ÷ 0.0123 ≈ 4.06 ≈ 4
So x = 4 and the empirical formula is:
CoCl₂·4H₂O
A sample contains 3.2×10⁻³ moles of helium gas. Convert this amount to the number of atoms using scientific notation and correct significant figures.
1.9×10²¹ atoms.
Number of atoms = (3.2×10⁻³ mol) × (6.022×10²³ atoms/mol)
= 1.93×10²¹ atoms
With two significant figures (from 3.2):
1.9×10²¹ atoms.
You have 1.00 mol of each compound: MgCl₂, K₂SO₄, and Al₂O₃. Which sample contains the greatest total number of ions? Explain your reasoning.
Al₂O₃.
MgCl₂ gives 3 ions per formula unit, K₂SO₄ gives 3 ions, and Al₂O₃ gives 5 ions (2 Al³⁺ and 3 O²⁻).
With 1.00 mol of each compound, Al₂O₃ produces the most ions.
A hydrate of magnesium sulfate has the formula MgSO₄·xH₂O. A student heats a sample and removes all the water. The percent composition of magnesium in the hydrate is measured to be lower than the percent composition of magnesium in anhydrous MgSO₄.
Explain why this must be true without doing any calculations.
The hydrate must have a lower percent of magnesium because the water increases the total mass but does not add any magnesium.
A compound contains 48.0% carbon, 8.00% hydrogen, and 44.0% oxygen by mass. What is its empirical formula?
Empirical formula: C₃H₆O₂.
Convert each percent to moles (using 100 g total):
C: 48.0 g ÷ 12.01 ≈ 4.00 mol
H: 8.00 g ÷ 1.008 ≈ 7.94 mol
O: 44.0 g ÷ 16.00 = 2.75 mol
Divide all by the smallest (2.75):
C ≈ 4.00 ÷ 2.75 ≈ 1.45
H ≈ 7.94 ÷ 2.75 ≈ 2.89
O = 1
These are not whole numbers, but each is very close to a multiple of 0.5.
Multiply all by 2:
C ≈ 2.90 → 3
H ≈ 5.78 → 6
O = 2
Empirical formula: C₃H₆O₂.
An industrial process uses a hydrate of barium chloride, BaCl₂·xH₂O. A 2.00 kg sample of the hydrate is dried completely in a large furnace. After heating, the mass of the anhydrous BaCl₂ is 1.59 kg.
Determine the value of x in BaCl₂·xH₂O. Show how you found x.
BaCl₂·3H₂O
Mass of hydrate = 2.00 kg
Mass of anhydrous BaCl₂ = 1.59 kg
Mass of water lost:
2.00 kg − 1.59 kg = 0.41 kg = 410 g
Moles of H₂O:
410 g ÷ 18.0 g/mol ≈ 22.8 mol
Molar mass of BaCl₂:
Ba: 137.33 g/mol
Cl₂: 70.90 g/mol
Total ≈ 208.23 g/mol
Moles of BaCl₂:
1.59 kg = 1590 g
1590 g ÷ 208.23 g/mol ≈ 7.64 mol
Mole ratio H₂O : BaCl₂:
22.8 ÷ 7.64 ≈ 2.98 ≈ 3
So x = 3
Hydrate formula: BaCl₂·3H₂O
A student measures a sample as 0.00450 kilograms of sodium chloride. Convert this mass to the total number of ions present in the sample. Use correct metric conversions, scientific notation, and significant figures.
9.28×10²² ions
Convert kg to g:
0.00450 kg = 4.50 g
Moles of NaCl:
M(NaCl) ≈ 58.44 g/mol
n = 4.50 g ÷ 58.44 g/mol ≈ 0.0770 mol
Formula units of NaCl:
0.0770 mol × 6.022×10²³ ≈ 4.64×10²² formula units
Each formula unit produces 2 ions (Na⁺ and Cl⁻):
Total ions = 2 × 4.64×10²² ≈ 9.28×10²² ions
With three significant figures:
9.28×10²² ions
You have three samples, each containing exactly 0.500 mol of molecules:
Sample A: CO₂
Sample B: C₂H₆
Sample C: N₂O₅
Which sample contains the greatest total number of atoms? Show your reasoning briefly.
Sample B: C₂H₆
Reason
Count atoms per molecule:
CO₂ has 3 atoms.
C₂H₆ has 8 atoms.
N₂O₅ has 7 atoms.
Two different compounds both contain only carbon and oxygen. One compound is 42.9% carbon by mass, and the other is 27.3% carbon by mass. Without doing any calculations, explain which compound must have the higher oxygen-to-carbon ratio in its formula and why.
The compound that is 27.3% carbon must have the higher oxygen-to-carbon ratio.
Reason
If carbon makes up a smaller fraction of the total mass, then proportionally more of the mass must come from oxygen. That can only happen if the formula contains more oxygen atoms per carbon atom compared to the compound with 42.9% carbon.
A compound contains only nitrogen and oxygen. Analysis shows it is 30.4% nitrogen and 69.6% oxygen by mass. What is the empirical formula?
Empirical formula: NO₂.
Assume 100 g:
N: 30.4 g ÷ 14.01 ≈ 2.17 mol
O: 69.6 g ÷ 16.00 = 4.35 mol
Divide by the smallest:
N: 2.17 ÷ 2.17 ≈ 1
O: 4.35 ÷ 2.17 ≈ 2
Empirical formula: NO₂.
Two students are trying to determine the percent by mass of water in the same copper(II) sulfate hydrate, CuSO₄·xH₂O.
• Student A heats the sample gently and stops as soon as the color changes from blue to white.
• Student B heats the sample strongly for a long time and accidentally causes a tiny bit of the solid salt to spatter out of the crucible.
Assume Student A’s sample is not fully dehydrated, and Student B’s sample has lost a small amount of CuSO₄ along with all the water.
Compared to the true percent by mass of water in CuSO₄·xH₂O, predict whether each student’s calculated percent water will be too high, too low, or about correct. Explain briefly for both students.
Student A-- too low
Calculated percent water: too low.
Reason: She stopped heating early, so some water is still in the “anhydrous” mass. That makes the mass loss (water lost) smaller than it should be, so her calculated percent water is lower than the true value.
Student B-- too high
Calculated percent water: too high.
Reason: He lost some CuSO₄ when it spattered out, but he treats all the mass loss as if it were water. That makes the mass loss (which he thinks is all water) larger than it should be, so his calculated percent water is higher than the true value.
A student has a sample of carbon dioxide with a mass of 7.50×10² mg. Calculate the total number of atoms in the sample.
Use scientific notation and correct significant figures throughout.
3.06×10²² atoms
Convert mg to g
7.50×10² mg = 750 mg = 0.750 g
Convert grams to moles
M(CO₂) ≈ 44.01 g/mol
n = 0.750 g ÷ 44.01 g/mol ≈ 0.0170 mol
Convert moles to molecules
0.0170 mol × 6.022×10²³ ≈ 1.02×10²² molecules
Convert molecules to atoms
Each CO₂ molecule has 3 atoms.
Total atoms = 3 × 1.02×10²² = 3.06×10²² atoms
With three significant figures:
3.06×10²² atoms
You have the following three samples:
Sample A: 16.0 g of O₂
Sample B: 16.0 g of CH₄
Sample C: 16.0 g of CO₂
Which sample contains the fewest molecules? Show enough reasoning to justify your choice.
CO₂ (Sample C).
Quick reasoning:
n = m / M
O₂: 16.0 g ÷ 32.0 g/mol = 0.500 mol
CH₄: 16.0 g ÷ 16.0 g/mol ≈ 1.00 mol
CO₂: 16.0 g ÷ 44.0 g/mol ≈ 0.364 mol
Fewer moles means fewer molecules, so CO₂ has the fewest molecules.
A certain family of compounds all have the same empirical formula, CH₂O. Three pure compounds are analyzed:
Compound A: CH₂O
Compound B: C₂H₄O₂
Compound C: C₃H₆O₃
You obtain a 1.00 g pure sample of each compound.
For which compound will that 1.00 g sample contain the greatest mass of carbon, or will they all have the same mass of carbon? Explain your reasoning without doing detailed calculations.
All three 1.00 g samples contain the same mass of carbon.
Reason
All three formulas are simple multiples of CH₂O. That means the ratio of carbon mass to total mass is the same in each compound, so their percent composition is identical. If the percent carbon is the same, then 1.00 g of each compound must contain the same mass of carbon.
A compound is 54.5% carbon, 9.10% hydrogen, and 36.4% oxygen by mass. The molar mass of the compound is about 132 g/mol.
Determine the molecular formula of the compound.
Molecular formula: C₆H₁₂O₃.
Assume 100.0 g of the compound:
C: 54.5 g ÷ 12.01 g/mol ≈ 4.54 mol
H: 9.10 g ÷ 1.008 g/mol ≈ 9.03 mol
O: 36.4 g ÷ 16.00 g/mol = 2.28 mol
Divide all by the smallest value, 2.28:
C: 4.54 ÷ 2.28 ≈ 1.99 ≈ 2
H: 9.03 ÷ 2.28 ≈ 3.96 ≈ 4
O: 2.28 ÷ 2.28 = 1
Empirical formula: C₂H₄O
Empirical formula mass ≈ (2 × 12.01) + (4 × 1.008) + (1 × 16.00) ≈ 44 g/mol
Molar mass given ≈ 132 g/mol
132 g/mol ÷ 44 g/mol = 3
So the molecular formula is 3 times the empirical formula:
Molecular formula: C₆H₁₂O₃.
A hydrate of magnesium sulfate has the formula MgSO₄·xH₂O. A 2.46 g sample of the hydrate is heated to constant mass, leaving 1.20 g of anhydrous MgSO₄. Determine the value of x in MgSO₄·xH₂O. Show how you found x.
MgSO₄·7H₂O.
Mass of water lost
2.46 g − 1.20 g = 1.26 g H₂O
Moles of MgSO₄
M(MgSO₄) ≈ 120.4 g/mol
n(MgSO₄) = 1.20 g ÷ 120.4 g/mol ≈ 0.00997 mol
Moles of H₂O
M(H₂O) ≈ 18.0 g/mol
n(H₂O) = 1.26 g ÷ 18.0 g/mol ≈ 0.0700 mol
Mole ratio H₂O : MgSO₄
0.0700 ÷ 0.00997 ≈ 7.02 ≈ 7
So x = 7
Hydrate formula: MgSO₄·7H₂O.
Mass of water lost
2.46 g − 1.20 g = 1.26 g H₂O
Moles of MgSO₄
M(MgSO₄) ≈ 120.4 g/mol
n(MgSO₄) = 1.20 g ÷ 120.4 g/mol ≈ 0.00997 mol
Moles of H₂O
M(H₂O) ≈ 18.0 g/mol
n(H₂O) = 1.26 g ÷ 18.0 g/mol ≈ 0.0700 mol
Mole ratio H₂O : MgSO₄
0.0700 ÷ 0.00997 ≈ 7.02 ≈ 7
So x = 7
Hydrate formula: MgSO₄·7H₂O.
A sealed container holds 4.25×10⁴ mL of neon gas at STP. Convert this volume to the total number of atoms of neon in the container. Use correct metric conversions, scientific notation, and significant figures.
1.14×10²⁴ atoms
Convert mL to L
4.25×10⁴ mL × (1 L / 1000 mL) = 4.25×10¹ L
Convert volume to moles at STP
22.4 L = 1 mol gas
n = (4.25×10¹ L) ÷ (22.4 L/mol)
n ≈ 1.90 mol
Convert moles to atoms
1.90 mol × 6.022×10²³ atoms/mol ≈ 1.14×10²⁴ atoms
With three significant figures:
1.14×10²⁴ atoms