Question

Oxygen forms a number of compounds with sulfur, many of which are quite reactive. One such compound is 57.20 % by mass S . Calculate the number of grams of oxygen present in a sample of this compound that contains 50.00 grams of S .

Answer 1

Answer is: number of grams of oxygen present in a sample is 37,41 g.
m(S) = 50 g.
mass percentage (S) = 57,20 % = 0,572.
mass percentage (O) = 100% - 57,20% 
mass percentage(O) = 42,8% = 0,428.
m(O) = ?.
make proportion 57,20% : 42,8% = 50 g : m(O)
57,2% · m(O) = 42,8% · 50 g.
m(O) = 42,8% · 50 g ÷ 57,2%.
m(O) = 37,41 g.

Answer 2

Answer:

37.41 grams of oxygen present in a sample of this compound that contains 50.00 grams of S.

Explanation:

Mass of Sulfur present in the sample = 50.00 g

Mass of oxygen present in the sample = m

Total mass of the sample of the compound =  = 50.00 g + m

Percentage of sulfur by mass in the compound = 57.20%

$\%(S)=\frac{\text{Mass of sulfur}}{\text{Total mass of the compound}}\times 100$

$57.20\%=\frac{50.00 g}{50.00g + m}\times 100$

m = 37.41 g

37.41 grams of oxygen present in a sample of this compound that contains 50.00 grams of S.