Question

You have a solution with 2.52 moles of isopropanol (C3H8O). The solution weighs 521 grams. What is the percent composition of isopropanol in the mixture?

Answer 1

The percent composition of isopropanol is 29%

Step-by-step explanation:

Pato Answer

Answer 2

The Percent composition of isopropanol in the mixture is 29.07 %

Step-by-step explanation:

Step 1: Data given

Number of moles isopropanol (C3H8O) = 2.52 moles

Mass of the solution = 521 grams

Molar mass of isopropanol (C3H8O) = 60.1 g/mol

Step 2: Calculate mass of isopropanol

Mass isopropanol = moles isopropanol * molar mass isopropanol

Mass isopropanol = 2.52 moles * 60.1 g/mol

Mass isopropanol = 151.45 grams

Step 3: Calculate the percent composition of isopropanol in the mixture

Percent composition of isopropanol = (mass isopropanol / total mass of mixture) * 100 %

Percent composition of isopropanol = (151.45 grams / 521 grams ) * 100 %

Percent composition of isopropanol = 29.07 %

The Percent composition of isopropanol in the mixture is 29.07 %