**Question:**

A 4 g sugar cube (Sucrose: C_{12}H_{22}O_{11}) is dissolved in a 350 ml teacup of 80 °C water. What is the percent composition by mass of the sugar solution?

Given: Density of water at 80 °C = 0.975 g/ml

**Definition**

Percent Composition by Mass is the mass of the solute divided by the mass of the solution (mass of the solute plus mass of the solvent), multiplied by 100.

**Step 1 - Determine mass of solute**

We were given the mass of the solute in the problem. The solute is the sugar cube.

mass_{solute} = 4 g of C_{12}H_{22}O_{11}

**Step 2 - Determine mass of solvent**

The solvent is the 80 °C water. Use the density of the water to find the mass.

density = mass/volume

mass = density x volume

mass = 0.975 g/ml x 350 ml

mass_{solvent} = 341.25 g

**Step 3 - Determine the total mass of the solution**

m_{solution} = m_{solute} + m_{solvent}

m_{solution} = 4 g + 341.25 g

m_{solution} = 345.25 g

**Step 4 - Determine percent composition by mass of the sugar solution.**

percent composition = (m_{solute} / m_{solution}) x 100

percent composition = ( 4 g / 345.25 g) x 100

percent composition = ( 0.0116 ) x 100

percent compostion = 1.16%

**Answer:**

The percent composition by mass of the sugar solution is 1.16%