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3 years ago

Find the empirical formula of the following compound: 0.77 mol of iron atoms combined with 1.0 mol of oxygen atoms.

Chemistry

1 answer:

eimsori [14]3 years ago

7 0

The empirical formula is Fe₃O₄.

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of Fe to O.

I like to summarize the calculations in a table.

Element Moles Ratio¹ ×3² Integers³

Fe 0.77 1 3 3

O 1.0 1.3 3.9 4

¹ To get the molar ratio, you divide each number of moles by the smallest number (0.77).

² If the ratio is not close to an integer, multiply by a number (in this case, 3) to get numbers that are close to integers.

³ Round off these numbers to integers (3 and 4).

The empirical formula is Fe₃O₄.

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Explanation :

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

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So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

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As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

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$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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