A 24.00 g sample contains 14.60 g Cl and 9.400 g B. What is the percent composition (by mass) of boron in this sample?

The percent composition of boron in this sample is 39.17 %.

The mass of chlorine = 14.60 g

The mass of Boron = 9.400g

Mass of the sample = 24.00 g

The percentage composition of boron can be calculated below:

% composition of Boron = 9.4 / 24 × 100

% composition of Boron = 940 / 24

% composition of Boron = 39.1666666667

% composition of Boron ≈ 39. 17%

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39.2 %

Explanation:

The following data were obtained from the question:

Mass of sample = 24 g

Mass of Cl = 14.6 g

Mass of B = 9.4 g

Percentage composition of boron =?

The percentage composition (by mass) of boron in the sample can be obtained as illustrated below:

Percentage composition of boron = mass of B /mass of sample × 100

Percentage composition of boron = 9.4/24 × 100

Percentage composition of boron = 39.2 %

Therefore, the percentage composition (by mass) of boron in the sample is 39.2 %

Step-by-step explanation:

3y - 6 = 18

3y = 24

y = 8