Question

What is the percent by weight of sodium in sodium sulfate (Na2SO4)? (The molar mass of Na = 22.99, S = 32.07, and O = 16.00.) 16.18% 22.58% 32.37% 45.05% 51.17%

Answer 1

The total mass would be 142.05 g/mol. Since sodium is 22.99 g/mol and there are 2 sodium atoms, it would be 45.98 g/mol. Divide 45.98 by 142.05 and you get 32.37%

Answer 2

Answer: The percentage by weight of sodium in sodium sulfate is 32.37 %.

Explanation:

In $Na_2SO_4$, there are 2 sodium atoms, 1 sulfur atom and 4 oxygen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

$\text{Mass percent of sodium}=\frac{\text{Mass of sodium}}{\text{Molar mass of sodium sulfate}}\times 100$

Mass of sodium = $(2\times 22.99)=45.98g$

Mass of sodium sulfate = $[(2\times 22.99)+32.07+(4\times 16)]=142.05g$

Putting values in above equation, we get:

$\text{Mass percent of sodium}=\frac{45.98g}{142.05}\times 100=32.37\%$

Hence, the mass percent of sodium in sodium sulfate is 32.37%.