Chemistry · General chemistry I · Worked example

Find mass from volume and density

A sample occupies $25.0 mL$ and has density $1.20 g / mL$ . What is its mass?

Write what is given and what is asked

The volume is $25.0 mL$ and the density is $1.20 g / mL$ . The question asks for a mass in grams. The density is taken to apply to this sample at the temperature it was measured.

V = 25.0 mL, ρ = 1.20 g/mL

Read the density as a factor

$1.20 g / mL$ means $1.20 g$ for every $1 mL$ . It can be written either way up, and only one way cancels the milliliters in $25.0 mL$ .

\frac{1.20 g}{1 mL} \frac{1 mL}{1.20 g}

Arrange the factor and cancel the units

Put $mL$ on the bottom, opposite the $mL$ in $25.0 mL$ . The milliliters cancel and grams are left, the unit the question asked for. On the board, select that cancellation yourself in the factor chain.

25.0 mL \times \frac{1.20 g}{1 mL}

Calculate the mass

$25.0 \times 1.20 = 30.0$ . The $1$ under the $1.20 g$ only says “per milliliter”. It is part of the density, not a measurement with one significant figure. Both measured values have three figures, so the mass is $30.0 g$ , with the trailing zero written.

m = 25.0 \times 1.20 g = 30.0 g

Check the size and the relation

The liquid is a little denser than water, which is about $1 g / mL$ , so $25 mL$ of it should weigh a little more than $25 g$ . $30.0 g$ fits. The relation $m = ρ V$ gives the same product, $(1.20 g / mL) (25.0 mL)$ $= 30.0 g$ .

Result

The mass is $30.0 g$ .

Your turn

1. One like the example

A $40.0 mL$ sample has a density of $0.850 g / mL$ . What is its mass?

Show the answer and explanation

$34.0 g$ .

Multiply $40.0 mL$ by $0.850 g$ over $1 mL$ , so the milliliters cancel and grams are left. $40.0 \times 0.850 = 34.0 g$ , three figures like both measurements. The density is below $1 g / mL$ , so the mass should be less than $40 g$ , and it is.

\begin{array}{l} V = 40.0 mL \\ ρ = 0.850 g/mL \\ m = ρ V \\ m = (0.850 g/mL) (40.0 mL) = 34.0 g \end{array}

\begin{array}{c} V = 40.0 mL \\ ρ = 0.850 g/mL \\ m = ρ V \\ m = (0.850 g/mL) (40.0 mL) \\ = 34.0 g \end{array}

2. A variation

Ethanol has a density of $0.789 g / mL$ . What volume does $50.0 g$ of ethanol occupy?

Show the answer and explanation

$63.4 mL$ .

Grams must cancel now, so the density goes in upside down, $1 mL$ over $0.789 g$ . $50.0 \div 0.789 = 63.37 mL$ , which is $63.4 mL$ to three figures. In relation form, $V = m / ρ$ . Ethanol is less dense than water, so its volume is more than $50 mL$ .

\begin{array}{l} m = 50.0 g \\ ρ = 0.789 g/mL \\ V = \frac{m}{ρ} \\ V = \frac{50.0 g}{0.789 g/mL} \\ V = 63.4 mL \end{array}

3. Somewhere new

A $2.00 L$ bottle is full of a liquid whose density is $1.26 g / mL$ . What is the mass of the liquid in kilograms?

Show the answer and explanation

$2.52 kg$ .

The density is per milliliter, so convert the volume first. $2.00 L \times 1000 mL / 1 L$ $= 2.00 \times 10^{3} mL$ . Then $m$ $= (1.26 g / mL) (2.00 \times 10^{3} mL)$ $= 2.52 \times 10^{3} g$ , and $1 kg / 1000 g$ turns that into $2.52 kg$ . The two metric factors are exact, so three figures remain.

\begin{array}{l} V = 2.00 L \\ 2.00 L \times \frac{1000 mL}{1 L} = 2.00 \times 10^{3} mL \\ ρ = 1.26 g/mL \\ m = ρ V \\ m = (1.26 g/mL) (2.00 \times 10^{3} mL) = 2.52 \times 10^{3} g \\ 2.52 \times 10^{3} g \times \frac{1 kg}{1000 g} = 2.52 kg \end{array}

\begin{array}{c} V = 2.00 L \\ 2.00 L \times \frac{1000 mL}{1 L} \\ = 2.00 \times 10^{3} mL \\ ρ = 1.26 g/mL \\ m = ρ V \\ m = (1.26 g/mL) (2.00 \times 10^{3} mL) \\ = 2.52 \times 10^{3} g \\ 2.52 \times 10^{3} g \times \frac{1 kg}{1000 g} \\ = 2.52 kg \end{array}

Keep exploring

Open the rows in a Chemistry box and change $30.0 g$ to $3.00 g$ . The last row turns ✗: “Arithmetic: this works out to 30, not 3.00.”

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 6 key terms Build the factor chain Check the mass in a Chemistry box Open worked example on a board Conversion factors

Your existing work stays on this device. Examples open as editable copies.

Write what is given and what is asked

Read the density as a factor

Arrange the factor and cancel the units

Calculate the mass

Check the size and the relation

Result

Your turn

1. One like the example

34.0 g34.0 g.

2. A variation

63.4 mL63.4 mL.

3. Somewhere new

2.52 kg2.52 kg.

Keep exploring

Study it, then try it

$34.0 g$ .

$63.4 mL$ .

$2.52 kg$ .