Percent Composition by Mass Example Problem

Percent composition by mass is a statement of the percent mass of each element in a chemical compound or the percent mass of components of a solution or alloy. This worked example chemistry problem works through the steps to calculate percent composition by mass. The example is for a sugar cube dissolved in a cup of water. Percent Composition by Mass Question A 4 g sugar cube (Sucrose: C12H22O11) is dissolved in a 350 ml teacup of 80 Â °C water. What is the percent composition by mass of the sugar solution? Given: Density of water at 80 Â °C 0.975 g/ml Percent Composition 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. How to Solve the Problem Step 1 - Determine mass of solute We were given the mass of the solute in the problem. The solute is the sugar cube. masssolute 4 g of C12H22O11 Step 2 - Determine mass of solvent The solvent is the 80 Â °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 masssolvent 341.25 g Step 3 - Determine the total mass of the solution msolution msolute msolvent msolution 4 g 341.25 g msolution 345.25 g Step 4 - Determine percent composition by mass of the sugar solution. percent composition (msolute / msolution) x 100 percent composition ( 4 g / 345.25 g) x 100 percent composition ( 0.0116) x 100 percent composition 1.16% Answer: The percent composition by mass of the sugar solution is 1.16% Tips for Success Its important to remember you use the total mass of the solution and not just the mass of the solvent. For dilute solutions, this doesnt make a huge difference, but for concentrated solutions, youll get a wrong answer.If youre given the mass of solute and mass of solvent, life is easy, but if youre working with volumes, youll need to use density to find the mass. Remember density varies according to temperature. Its unlikely youll find a density value corresponding to your exact temperature, so expect this calculation to introduce a small amount of error into your calculation.