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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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Sati [7]

Answer:

  1. C-B
  2. C-C
  3. C-N
  4. C-O
  5. C-F

Explanation:

As we move along to the <u>right in the same period, the electronegativity</u> and <u>the effective nuclear charge values are higher.</u>

The tendency is that <em>the higher these values are, the shorter the bonds will be</em>.

With that information in mind, and looking at the periodic table, the order would be:

  1. C-B
  2. C-C
  3. C-N
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  5. C-F

Where the C-F bond is the shortest among them.

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A sample of quartz is put into a calorimeter (see sketch at right) that contains of water. The quartz sample starts off at and t
pashok25 [27]

Answer:

0.71 J/g°C

Explanation:

Here is the complete question

thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C

Solution

Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.

So, heat lost by quartz, Q = heat gained by water, Q'

-Q = Q'

-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C

Making c subject of the formula, we have

c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)

Substituting the values of the variables into the equation, we have

c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)

c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)

c = -2898 J/-4074.15 g°C

c = 0.711 J/g°C

c ≅ 0.71 J/g°C to 2 significant digits

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