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

`Sn_2O`

Step-by-step 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=(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*(1molSn)/(118.8gSn) =0.788molSn\\\\n_O=6.4gO*(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:(0.788)/(0.4)=2\\ \\O:(0.4)/(0.4)=1`

Therefore, the empirical formula is:

`Sn_2O`

Best regards.