Percent by Weight Calculation

percent by weight calculation solute solvent solution

Core Concepts

In this tutorial, you will be introduced to a calculation called percent by weight. In order to fully understand, you will also walk through an example.

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What is Percent by Weight?

Percent by weight, also referred to as mass percent composition, can be abbreviated as w/w\%. This is a type of calculation that allows you to find the ratio of solute to solution. You can use the equation below to help solve example problems:

    \begin{align*} {\% \text{by Weight} = \frac{\text{g of solute}}{\text{g of solute} + \text{g of solvent}} \left(x100\right)} \end{align*}

solvent solute solution percent by weight calculation

How To Calculate Percent By Weight

When doing this calculation, the way you arrive at your answer depends on the information given to you. However, each method ends the same way: you need to find the ratio of how many grams of solute there are for the total grams of solution.

First, when doing any calculation, you have to make sure you are using consistent units. Sometimes the amounts in grams will be given, but sometimes you might have to do some extra work to get it to the same units. For example, if an amount is given in milligrams and the other in kilograms, you will have to convert using the correct conversions. You may also have an amount in moles, which you can convert by using the molar mass. From now on, we will refer to the units in grams to stay consistent.

Then, you have to find the total grams of solution, which includes the solvent and solute. You should already have the grams of solute, so you can use the same methods to find the grams of solvent.

Lastly, you plug the numbers into the percent by weight equation and get your answer.

Let’s walk through an example below to further understand this calculation!

Percent by Weight Example

Determine the percent by weight of sodium chloride in the following solution: 1.75\text{moles} of NaCl (58.44\text{g/mol}) dissolved in 0.550\text{kg} of deionized H2O.

To start this problem, we need to convert our information into the same unit, in this case, grams. Using molar masses and unit conversions, we are able to find the grams of solute (NaCl) and solvent (Water).

    \begin{align*} {1.75\text{mol}NaCl \cdot \frac{58.44\text{g}NaCl}{1\text{mol}NaCl} &= 102.27\text{g}NaCl} \\ {0.550\text{kg} H_{2}O \cdot \frac{1000\text{g}H_{2}O}{1\text{kg}H_{2}O} &= 550\text{g}H_{2}O} \end{align*}

Since we have all the information we need, the final step is to plug everything into our formula. By doing this, we see that this solution is 15.68\% NaCl.

    \begin{align*} {\% \text{by Weight} = \frac{\text{g of solute}}{\text{g of solute} + \text{g of solvent}} = \frac{102.27\text{g}}{102.27\text{g} + 550\text{g}} = \frac{102.27\text{g}}{652.27\text{g}} = 0.1568\left(x100\right) = 15.68\%} \end{align*}

Percent By Weight Practice Problems

Problem 1

Calculate the percent by weight of acetic acid (CH3COOH) (60.05\text{g/mol}) in a 1.00\text{M} aqueous solution. Assume water has a density of 1.00\text{g/mL}.

Problem 2

From a 10\% NaOH solution, calculate the expected weight of NaOH when you evaporate the water out of 54\text{mL}.

Percent By Weight Practice Problem Solutions

1: 6.01\% acetic acid

2: 5.4\text{g} NaOH

Further Reading