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? round your answer to the nearest whole percent. use the periodic table to help you find the necessary molar masses. select the correct option from the drop - down menu. the percent composition of isopropanol is %.

Explanation:

Step1: Calculate molar mass of isopropanol ($\ce{C3H8O}$)

Molar mass of C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.
Molar mass = \( 3\times12.01 + 8\times1.008 + 16.00 \)
= \( 36.03 + 8.064 + 16.00 \)
= \( 60.094 \) g/mol.

Step2: Calculate mass of isopropanol

Moles of isopropanol = 2.52 mol.
Mass = moles × molar mass = \( 2.52 \times 60.094 \)
≈ \( 151.43688 \) g.

Step3: Calculate percent composition

Percent composition = \( \frac{\text{Mass of isopropanol}}{\text{Mass of solution}} \times 100 \)
= \( \frac{151.43688}{521} \times 100 \)
≈ \( 29.0666\% \), rounded to nearest whole percent is 29%.

Answer:

29%