Answer
The mass of S in 40.8 g $CaSO_{4}$ is 9.60 g.
The mass of S in 38.8 g $Na_{2}SO_{3}$ is 9.85 g.
Therefore 38.8 g $Na_{2}SO_{3}$ contains more S than 40.8 g $CaSO_{4}$.
Work Step by Step
Strategy: To find which of the given amounts of compaunds contains more S : 1- find the Molar mass of each compound 2- Calculate the percentage of S in each 3 - Calculate the amount of S in the given amount of each compound 4 - Compare the amounts
Calcium sulfate $CaSO_{4}$.
Molar mass $CaSO_{4}$ = 1 x 40.0 + 1 x 32.0 + 4 x 16.0
Molar mass $CaSO_{4}$ = 136 g/mol
Mass percentage S ( in $CaSO_{4}$) = $\frac{mass_(1 mole S)}{Molar mass_(CaSO_{4})}$$\times$ 100%
Mass percentage S ( in $CaSO_{4}$) = $\frac{mass_(1\times32.0)}{136}$$\times$ 100%
Mass percentage S ( in $CaSO_{4}$) = 23.5%
The mass of S in 40.8 g in $CaSO_{4}$ is 23.5% of 40.8:
23.5%$\times$40.8g = $\frac{23.5}{100}\times40.8g$
The mass of S in 40.8 g $CaSO_{4}$ is 9.60 g.
Sodium sulfite $Na_{2}SO_{3}$.
Molar mass $Na_{2}SO_{3}$ = 2 x 23.0 + 1 x 32.0 + 3x 16.0
Molar mass $Na_{2}SO_{3}$ = 126 g/mol
Mass percentage S ( in $Na_{2}SO_{3}$) = $\frac{mass_(1 mole S)}{Molar mass_(Na_{2}SO_{3})}$$\times$ 100%
Mass percentage S ( in $Na_{2}SO_{3}$) = $\frac{mass_(1\times32.0)}{126}$$\times$ 100%
Mass percentage S ( in $Na_{2}SO_{3}$) = 25.4%
The mass of S in 38.8 g in $Na_{2}SO_{3}$ is 25.4% of 38.8:
25.4%$\times$38.8g = $\frac{25.4}{100}\times38.8g$
The mass of S in 38.8 g $Na_{2}SO_{3}$ is 9.85 g.
Therefore 38.8 g $Na_{2}SO_{3}$ contains more S than 40.8 g $CaSO_{4}$.
