Question

A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?

Answer 1

Answer:

$Sn_2O$

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

$\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%$

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

$n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO$

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

$Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1$

Therefore, the empirical formula is:

$Sn_2O$

Best regards.