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kogti [31]
3 years ago
7

A compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula?

Chemistry
2 answers:
Ahat [919]3 years ago
6 0

Answer:

                           O        H        C

Moles in 100g  3.33   6.65    3.33

Ratio                  1.00   2.00    1.00

Possible empirical formula = OH_{2}C

Oduvanchick [21]3 years ago
4 0

Answer:

CH2O

Explanation:

Data obtained from the question include:

C = 40%

H = 6.7%

O = 53.3%

Divide by their molar mass

C = 40/12 = 3.33

H = 6.7/1 = 6.7

O = 53.3/16 = 3.33

Divide by the smallest

C = 3.33/3.33 = 1

H = 6.7/3.33 = 2

O = 3.33/3.33 = 1

The empirical formula is CH2O

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Identify the precipitate that forms (if any) when aqueous solutions of barium nitrate and sodium sulfate are mixed.
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4 0
1 year ago
Help?
Ahat [919]

Answer:

KClO_3

Explanation:

Hello!

In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:

m_O=22.9g-7.33g-6.65g=8.92g

Next, we compute the moles of each element:

n_K=\frac{7.33g}{39.9g/mol}= 0.184mol\\\\n_{Cl}=\frac{6.65g}{35.45g/mol}=0.188mol \\\\n_O=\frac{8.92g}{16.00g/mol} =0.5575mol

Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:

K=\frac{0.184}{0.184}=1.0 \\\\Cl=\frac{0.187}{0.184}=1.0\\\\O=\frac{0.5575}{0.184}  =3.0

Therefore, the empirical formula is:

KClO_3

Regards!

3 0
2 years ago
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