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enyata [817]
3 years ago
5

A compound containing only c, h, and o, was extracted from the bark of the sassafras tree. The combustion of 43.9 mg produced 11

9 mg of co2 and 24.4 mg of h2o. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.
Chemistry
1 answer:
katen-ka-za [31]3 years ago
3 0

The compound contain only carbon, hydrogen and oxygen, let the empirical formula of compound be C_{x}H_{y}O_{z}.

The mass of compound, carbon dioxide and water is 43.9 mg, 119 mg and 24.4 mg respectively . The molar mass of compound, carbon dioxide and water is 162 g/mol, 44.01 g/mol and 18 g/mol respectively.

Converting the mass into number of moles as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Number of moles of carbon dioxide CO_{2} will be:

n=\frac{119\times 10^{-3}}{44.01 g/mol }=0.0027 mol

1 mol of CO_{2} has 1 mol of C thus, number of moles of C from 0.0027 mol of CO_{2} will be 0.0027 mol.

Molar mass of C is 12 g/mol thus, mass of C will be:

m_{C}=0.0027 mol\times 12 g/mol=0.0324 g

Number of moles of water H_{2}O will be:

n=\frac{24.4\times 10^{-3}}{18 g/mol }=0.001356 mol

1 mol of H_{2}O has 2 mol of H  thus, 0.001356 mol of H_{2}O will have 2\times 0.001356 mol=0.002712 mol mole of H.

Molar mass of H is 1 g/mol thus, mass of H will be:

m_{H}=0.002712 mol\times 1 g/mol=0.002712 g

Sum of mass of C and H will be:

m_{C+H}=0.0324+0.002712=0.035112 g

Mass of compound is 43.9 mg or 0.0439 g thus, mass of oxygen will be:

m_{O}=m_{compound}-m_{C+H}=(0.0439-0.035112)g=0.008788 g

Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:

n=\frac{0.008788 g}{16 g/mol}=0.00055 mol

Taking the molar ratio:

C:H:O=0.0027:0.002712:0.00055=5:5:1

Therefore, empirical formula will be C_{5}H_{5}O.

The molar mass of C_{5}H_{5}O is 81 g/mol

The molar mass of given compound is 162 that is 2 times the molar mass calculated above thus, molecular formula will be 2\times C_{5}H_{5}O=C_{10}H_{10}O_{2}.

Therefore, empirical formula and molecular formula of the compound is C_{5}H_{5}O and C_{10}H_{10}O_{2} respectively.



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You can calculate the concentration of hydronium ions using antilogarithm properties:

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What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH
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1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2:    M(H2) = 2*1.0= 2.0 g/mol
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<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover. 

                                  CO       +         2H2   ------->      CH3OH  
                                 1 mol              2 mol
given                        2.66 mol          6 mol (excess)

How much
we need  CO?           3 mol              6 mol

We see that H2 will be leftover, because for 6 moles H2  we need 3 moles CO, but we have only 2.66 mol  CO.
So, CO will react completely, and we are going to use CO to find  the mass of CH3OH.

3)                              </span>CO       +         2H2   ------->      CH3OH  
                                 1 mol                                        1 mol
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4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 +  4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH =  85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
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The statement above suggests that 1 mole of chromium contains 6.02x10^23 atoms.

Now, if 1mole of of chromium contains 6.02x10^23 atoms,

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Xmol of chromium = 7.52x10^22/6.02x10^23

Xmol of chromium = 0.125 mole

Therefore, 0.125 mole of chromium contains 7.52x10^22 atoms

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