Question

A 12.41 g sample of nabr contains 22.34% na by mass. considering the law of constant composition (definite proportions) how many grams of sodium does a 8.99 g sample of sodium bromide contain?

Answer 1

Answer:

$m_{Na}=2.01g$

Explanation:

Hello,

In this case, given the by mass percent of sodium in sodium bromide::

$\% Na=\frac{m_{Na}}{m_{NaBr}}*100\%=22.34 \%$

Solving for the mass of sodium, considering the constant composition of sodium in the sodium bromide, we obtain:

$m_{Na}=\frac{\% Na*m_{NaBr}}{100\%} =\frac{22.34\%*8.99g}{100\%} \\\\m_{Na}=2.01g$

Best regards.

Answer 2

I'm assuming you mean how many grams of sodium does 8.99 grams of sodium bromide have, because Bromine would not have any sodium in it.

Then, just direct calculate 22.34
$(22.34 \div 100) \times 8.99 = 2.008366$