Percentages Quiz - Complete Answer Key

Detailed Solutions with Step-by-Step Work

Section A Basic Percentage Calculations

Problem 1

A quarter from 1964 is made of 90% silver and has a mass of 5.670 g. What is the mass of silver in the quarter?

Step 1: Identify the WHOLE

The WHOLE is the total mass of the quarter = 5.670 g

Step 2: Convert percentage to decimal

\[90\% = 0.90\]

Step 3: Calculate Part = Decimal × Whole

\[\text{Mass of silver} = 0.90 \times 5.670 \text{ g}\]

\[= 5.103 \text{ g} = 5.10 \text{ g (3 sig figs)}\]

Answer: 5.10 g of silver

Problem 2

A sample of brass contains 65% copper. If the brass sample has a mass of 250.0 g, how many grams of copper does it contain?

Step 1: Convert percentage to decimal

\[65\% = 0.65\]

Step 2: Calculate

\[\text{Mass of Cu} = 0.65 \times 250.0 \text{ g} = 162.5 \text{ g}\]

With sig figs: 160 g (2 sig figs from 65%)

Answer: 160 g of copper (or 162.5 g if treating 65% as exact)

Problem 3

Pure gold is 24 karat. If you have an 18-karat gold ring (75% gold) with a mass of 8.50 g, what is the mass of pure gold in the ring?

Step 1: Convert percentage to decimal

\[75\% = 0.75\]

Step 2: Calculate

\[\text{Mass of gold} = 0.75 \times 8.50 \text{ g}\]

\[= 6.375 \text{ g} = 6.4 \text{ g (2 sig figs)}\]

Answer: 6.4 g of pure gold

Problem 4

A 500.0 mL sample of seawater contains 3.5% dissolved salts by mass. If the density of seawater is 1.025 g/mL, what mass of salt is present?

Step 1: Find mass of seawater

\[\text{Mass} = \text{Volume} \times \text{Density}\]

\[= 500.0 \cancel{\text{ mL}} \times \frac{1.025 \text{ g}}{1 \cancel{\text{ mL}}} = 512.5 \text{ g}\]

Step 2: Find mass of salt (3.5% of total mass)

\[3.5\% = 0.035\]

\[\text{Mass of salt} = 0.035 \times 512.5 \text{ g}\]

\[= 17.9375 \text{ g} = 18 \text{ g (2 sig figs)}\]

Answer: 18 g of salt

Section B Atmospheric & Gas Composition

Problem 5

The atmosphere is 20.95% oxygen. If you breathe in 0.5 L of air, what volume of oxygen did you breathe?

Step 1: Convert percentage to decimal

\[20.95\% = 0.2095\]

Step 2: Calculate volume of O₂

\[\text{Volume of O}_2 = 0.2095 \times 0.5 \text{ L}\]

\[= 0.10475 \text{ L} = 0.10 \text{ L (1 sig fig)}\]

Answer: 0.10 L of oxygen

Problem 6

Earth's atmosphere is 78.08% nitrogen by volume. A room contains 450 m³ of air. What volume of nitrogen is in the room?

Step 1: Convert percentage to decimal

\[78.08\% = 0.7808\]

Step 2: Calculate volume of N₂

\[\text{Volume of N}_2 = 0.7808 \times 450 \text{ m}^3\]

\[= 351.36 \text{ m}^3 = 350 \text{ m}^3 \text{ (2 sig figs)}\]

Answer: 350 m³ of nitrogen

Problem 7

Carbon dioxide makes up 0.04% of the atmosphere. If you breathe 12 L of air, how many liters of CO₂ did you inhale?

Step 1: Convert percentage to decimal

\[0.04\% = 0.0004\]

Step 2: Calculate volume of CO₂

\[\text{Volume of CO}_2 = 0.0004 \times 12 \text{ L}\]

\[= 0.0048 \text{ L} = 0.005 \text{ L} = 5 \text{ mL}\]

Answer: 0.005 L or 5 mL of CO₂

Problem 8

A helium-oxygen mixture used in deep-sea diving contains 96% helium and 4% oxygen. If a diver's tank holds 3.5 L of this mixture, how many liters of oxygen are in the tank?

Step 1: Convert percentage to decimal

\[4\% = 0.04\]

Step 2: Calculate volume of O₂

\[\text{Volume of O}_2 = 0.04 \times 3.5 \text{ L}\]

\[= 0.14 \text{ L}\]

Answer: 0.14 L of oxygen

Section C Atomic & Molecular Calculations

Problem 9

Your body contains about 16 kg of carbon, and 18% of that carbon is carbon-12. If you have \(3.0 \times 10^{27}\) carbon atoms total, how many are carbon-12?

Step 1: Convert percentage to decimal

\[18\% = 0.18\]

Step 2: Calculate number of C-12 atoms

\[\text{C-12 atoms} = 0.18 \times 3.0 \times 10^{27}\]

\[= 0.54 \times 10^{27} = 5.4 \times 10^{26} \text{ atoms}\]

Answer: \(5.4 \times 10^{26}\) carbon-12 atoms

Problem 10

A sample of chlorine gas contains 75.78% Cl-35 and 24.22% Cl-37. If the sample contains \(1.5 \times 10^{24}\) total chlorine atoms, how many Cl-35 atoms are present?

Step 1: Convert percentage to decimal

\[75.78\% = 0.7578\]

Step 2: Calculate number of Cl-35 atoms

\[\text{Cl-35 atoms} = 0.7578 \times 1.5 \times 10^{24}\]

\[= 1.1367 \times 10^{24} = 1.1 \times 10^{24} \text{ atoms}\]

Answer: \(1.1 \times 10^{24}\) Cl-35 atoms

Problem 11

A bronze statue is made of 88% copper and 12% tin. If the statue contains \(2.5 \times 10^{26}\) total metal atoms, how many are copper atoms?

Step 1: Convert percentage to decimal

\[88\% = 0.88\]

Step 2: Calculate number of Cu atoms

\[\text{Cu atoms} = 0.88 \times 2.5 \times 10^{26}\]

\[= 2.2 \times 10^{26} \text{ atoms}\]

Answer: \(2.2 \times 10^{26}\) copper atoms

Problem 12

Natural boron consists of 19.9% B-10 and 80.1% B-11. A sample contains \(6.02 \times 10^{23}\) boron atoms (1 mole). How many B-11 atoms are in the sample?

Step 1: Convert percentage to decimal

\[80.1\% = 0.801\]

Step 2: Calculate number of B-11 atoms

\[\text{B-11 atoms} = 0.801 \times 6.02 \times 10^{23}\]

\[= 4.82 \times 10^{23} \text{ atoms}\]

Answer: \(4.82 \times 10^{23}\) B-11 atoms

Section D Multi-Step Value Problems

Problem 13

If silver is worth $31.02 per troy ounce, and there are 31.1 g in 1 troy ounce, what is the value of silver in all 1964 quarters ever minted (1,950,000,000 quarters)?

Step 1: Find silver per quarter (from Problem 1)

Each quarter: 0.90 × 5.670 g = 5.10 g of silver

Step 2: Find total silver mass

\[\text{Total silver} = 5.10 \text{ g/quarter} \times 1.95 \times 10^9 \text{ quarters}\]

\[= 9.945 \times 10^9 \text{ g}\]

Step 3: Convert to troy ounces

\[9.945 \times 10^9 \cancel{\text{ g}} \times \frac{1 \text{ troy oz}}{31.1 \cancel{\text{ g}}} = 3.197 \times 10^8 \text{ troy oz}\]

Step 4: Calculate dollar value

\[3.197 \times 10^8 \cancel{\text{ troy oz}} \times \frac{\$31.02}{1 \cancel{\text{ troy oz}}} = \$9.92 \times 10^9\]

Answer: $9.92 billion (or $9,920,000,000)

Problem 14

A copper mine produces ore that is 0.75% copper by mass. If copper sells for $4.50 per pound and there are 453.6 g in 1 pound, what is the value of copper in 1000 kg of ore?

Step 1: Find mass of copper in ore

\[0.75\% = 0.0075\]

\[\text{Mass of Cu} = 0.0075 \times 1000 \text{ kg} = 7.5 \text{ kg}\]

Step 2: Convert to grams

\[7.5 \cancel{\text{ kg}} \times \frac{1000 \text{ g}}{1 \cancel{\text{ kg}}} = 7500 \text{ g}\]

Step 3: Convert to pounds

\[7500 \cancel{\text{ g}} \times \frac{1 \text{ lb}}{453.6 \cancel{\text{ g}}} = 16.53 \text{ lb}\]

Step 4: Calculate dollar value

\[16.53 \cancel{\text{ lb}} \times \frac{\$4.50}{1 \cancel{\text{ lb}}} = \$74.39 \approx \$74\]

Answer: $74

Problem 15

A sterling silver necklace is 92.5% silver and has a mass of 45.0 g. If silver is worth $0.85 per gram, what is the silver value of the necklace?

Step 1: Find mass of pure silver

\[92.5\% = 0.925\]

\[\text{Mass of Ag} = 0.925 \times 45.0 \text{ g} = 41.625 \text{ g}\]

Step 2: Calculate dollar value

\[41.625 \cancel{\text{ g}} \times \frac{\$0.85}{1 \cancel{\text{ g}}} = \$35.38 \approx \$35\]

Answer: $35

Section E Percent Composition Challenges

Problem 16

A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. If you have 250.0 g of the compound, how many grams of each element are present?

Carbon:

\[0.400 \times 250.0 \text{ g} = 100.0 \text{ g C}\]

Hydrogen:

\[0.067 \times 250.0 \text{ g} = 16.75 \approx 17 \text{ g H}\]

Oxygen:

\[0.533 \times 250.0 \text{ g} = 133.25 \approx 133 \text{ g O}\]

Answer: 100.0 g C, 17 g H, 133 g O

Problem 17

Ammonia (NH₃) is 82.35% nitrogen by mass. How many grams of nitrogen are in 150.0 g of ammonia?

Step 1: Convert percentage to decimal

\[82.35\% = 0.8235\]

Step 2: Calculate mass of nitrogen

\[\text{Mass of N} = 0.8235 \times 150.0 \text{ g}\]

\[= 123.525 \text{ g} = 123.5 \text{ g}\]

Answer: 123.5 g of nitrogen

Problem 18

A fertilizer contains 15% nitrogen, 30% phosphorus, and 15% potassium (15-30-15 fertilizer). A bag contains 25.0 kg of fertilizer.

Part a: Mass of nitrogen

\[0.15 \times 25.0 \text{ kg} = 3.75 \text{ kg N}\]

Part b: Mass of phosphorus

\[0.30 \times 25.0 \text{ kg} = 7.5 \text{ kg P}\]

Part c: Percentage that is NOT N, P, or K

\[\text{Total NPK} = 15\% + 30\% + 15\% = 60\%\]

\[\text{Other} = 100\% - 60\% = 40\%\]

Answer: a. 3.75 kg N, b. 7.5 kg P, c. 40%

Problem 19

A sample of air in Los Angeles contains 0.15% carbon monoxide by volume (a dangerously high level!). If you breathe 8.5 L of this air, what volume of toxic CO did you inhale?

Step 1: Convert percentage to decimal

\[0.15\% = 0.0015\]

Step 2: Calculate volume of CO

\[\text{Volume of CO} = 0.0015 \times 8.5 \text{ L}\]

\[= 0.01275 \text{ L} = 0.013 \text{ L} = 13 \text{ mL}\]

Answer: 0.013 L or 13 mL of CO

Problem 20

The human body is approximately 65% oxygen by mass. If a person weighs 70.0 kg, what mass of oxygen is in their body? How does this compare to the 16 kg of carbon in a typical human body?

Step 1: Calculate mass of oxygen

\[65\% = 0.65\]

\[\text{Mass of O} = 0.65 \times 70.0 \text{ kg} = 45.5 \text{ kg}\]

Step 2: Compare to carbon

\[\frac{45.5 \text{ kg O}}{16 \text{ kg C}} = 2.84\]

The body contains about 2.8 times more oxygen than carbon by mass.

Answer: 45.5 kg of oxygen; the body has about 2.8× more oxygen than carbon