Density Calculations Worksheet I - Free Printable

Let's solve each problem step by step using the density formula:

$$
\text{Density (D)} = \frac{\text{Mass (m)}}{\text{Volume (v)}}
$$

And its rearranged forms:
- $ m = D \times v $
- $ v = \frac{m}{D} $

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1. Find the unknown quantity for a, b, and c:

#### a) D = 3 g/mL, v = 100 mL, m = ? g

Use:
$$
m = D \times v = 3 \, \text{g/mL} \times 100 \, \text{mL} = 300 \, \text{g}
$$

✔ Answer: 300 g

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#### b) D = ? g/mL, v = 950 mL, m = 95 g

Use:
$$
D = \frac{m}{v} = \frac{95 \, \text{g}}{950 \, \text{mL}} = 0.1 \, \text{g/mL}
$$

✔ Answer: 0.1 g/mL

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#### c) D = 0.5 g/cm³, v = ? mL, m = 20 g

Note: Since density is in g/cm³, and volume is asked in mL, recall that 1 cm³ = 1 mL, so units are equivalent.

Use:
$$
v = \frac{m}{D} = \frac{20 \, \text{g}}{0.5 \, \text{g/cm³}} = 40 \, \text{cm³} = 40 \, \text{mL}
$$

✔ Answer: 40 mL

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2. Find the unknown quantity:

#### a) D = 24 g/mL, v = 1200 mL, m = ? g

$$
m = D \times v = 24 \, \text{g/mL} \times 1200 \, \text{mL} = 28,800 \, \text{g}
$$

✔ Answer: 28,800 g

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#### b) D = ? g/mL, v = 100 mL, m = 1500 g

$$
D = \frac{m}{v} = \frac{1500 \, \text{g}}{100 \, \text{mL}} = 15 \, \text{g/mL}
$$

✔ Answer: 15 g/mL

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#### c) D = ? g/mL, v = 520 mL, m = 500 g

$$
D = \frac{m}{v} = \frac{500 \, \text{g}}{520 \, \text{mL}} \approx 0.9615 \, \text{g/mL}
$$

Round to appropriate significant figures:
500 has 1 sig fig? Actually, it’s ambiguous, but likely 3 sig figs (since written as 500.0 would be clearer). But here, 500 g has 1 or 3 sig figs? Usually if no decimal, it's considered to have 1 sig fig unless specified. But in context, assume 3 sig figs.

So:
$$
D = \frac{500}{520} = 0.9615 \rightarrow \boxed{0.962 \, \text{g/mL}} \quad \text{(3 sig figs)}
$$

✔ Answer: 0.962 g/mL

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WORD PROBLEMS:

#### 1. A block of Aluminum occupies a volume of 15.0 mL and has a mass of 40.5 g. What is its density?

Formula:
$$
D = \frac{m}{v}
$$

Substitutions with units:
$$
D = \frac{40.5 \, \text{g}}{15.0 \, \text{mL}}
$$

Solution with units:
$$
D = \frac{40.5}{15.0} = 2.70 \, \text{g/mL}
$$

✔ Answer: 2.70 g/mL

(Note: 3 significant figures — both numbers have 3 sig figs.)

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#### 2. Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used to fill the cylinder has a mass of 306.0 g. Calculate the density of mercury.

Formula:
$$
D = \frac{m}{v}
$$

Substitutions with units:
$$
D = \frac{306.0 \, \text{g}}{22.5 \, \text{mL}}
$$

Solution with units:
$$
D = \frac{306.0}{22.5} = 13.6 \, \text{g/mL}
$$

✔ Answer: 13.6 g/mL

(3 significant figures: 306.0 has 4, 22.5 has 3 → limit to 3 sig figs)

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✔ Final Answers Summary:

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#### Part 1:
a) m = 300 g
b) D = 0.1 g/mL
c) v = 40 mL

#### Part 2:
a) m = 28,800 g
b) D = 15 g/mL
c) D = 0.962 g/mL

#### Word Problems:
1. Density of aluminum = 2.70 g/mL
2. Density of mercury = 13.6 g/mL

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