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 the mass of the sample and mass of tin we can compute the mass of oxygen via:

$m_O=0.534g-0.500g=0.034g$

Thus, by using the atomic bas of tin and oxygen we can compute their moles:

$n_{Sn}=0.500gSn*\frac{1molSn}{118.8gSn} =0.00421mol\\\\n_O=0.034gO*\frac{1molO}{16gO}=0.002125mol$

Next, we need to divide both moles by the moles of oxygen as those are the smallest in order to compute the subscript in the chemical reaction:

$Sn=\frac{0.00421}{0.002125}=2\\ \\O=\frac{0.002125}{0.002125}= 1$

Therefore, empirical formula of the oxide should be:

$Sn_2O$

Best regards.