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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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The mass percent of hydrogen in CH₄O is 12.5%.

<h3>What is the mass percent?</h3>

Mass percent is the mass of the element divided by the mass of the compound or solute.

  • Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

  • Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

  • Step 3: Calculate the mass percent of hydrogen in the compound.

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The mass percent of hydrogen in CH₄O is 12.5%.

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What is the poH of a<br> 2.6 x 10-6 M H+ solution?
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Explanation:

pH and pOH.....

The pH is a way of expressing the hydrogen ion concentration.

pH = -log[H+] ............. where [x] means "the concentration of x in moles per liter."

From pH you can compute pOH since at 25C pH + pOH = 14.00 .......... (but only at 25C)

pH = -log(2.6x10^-6) = 5.585 ..... which should be rounded to two significant digits: pH = 5.59

When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number. Since 2.6x10^-6 has two significant digits, a pH of 5.59 has two significant digits.

pOH + pH = 14.00

pOH = 14.00 - pH = 14.00 - 5.59 = 8.41 ......... at 25C

We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

Kw = [H+][OH-] = 1.00x10^-14 .......... at 25C .... like any Kc, the value changes with temperature

[OH-] = Kw / [H+] = 1.00x10^-14 / 2.6x10^-6 = 3.846x10^-9 .... to a couple of guard digits

pOH = -log[OH-] = -log(3.846x10^-9) = 8.415 ...... round to two significant digits: pOH = 8.42 ..... at 25C

=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

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