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dybincka [34]
3 years ago
15

What is the empirical formula of a compound that is 18.8% Li, 16.3% C , and 64.9% O

Chemistry
1 answer:
zlopas [31]3 years ago
3 0

Greetings!

To find the empirical formula you need the relative atomic mass of each element!

Li = 6.9

C = 12

O = 16

You can simply change the percentages into full grams

Li = 18.8g

C = 16.3g

O = 64.9


Then you use this to find the Number of moles = amount in grams / atomic mass

Li = 18.8 ÷ 6.9 = 2.7246

C = 16.3 ÷ 12 = 1.3583

O = 64.9 ÷ 16 = 4.0562

Then divide each number of moles by the smallest value:

Li = 2.7246 ÷ 1.3583 = 2.0

C = 1.3583 ÷ 1.3583 = 1

O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3


So that means that there are 2 Li, 1 C, and 3 O

Empirical formula would be:

Li₂CO₃


Hope this helps!

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How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
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</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
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<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
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<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

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