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 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*(1molSn)/(118.8gSn) =0.00421mol\\\\n_O=0.034gO*(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=(0.00421)/(0.002125)=2\\ \\O=(0.002125)/(0.002125)= 1`

Therefore, empirical formula of the oxide should be:

`Sn_2O`

Best regards.