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Elodia [21]
2 years ago
12

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?

Chemistry
1 answer:
larisa [96]2 years ago
5 0

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.

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A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233g of HCl and 0.403g of H2O. Determine the empirical
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Explanation:

For the reaction:

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<em></em>

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