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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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Which element is used to reduce metal oxides and make pure metals?
Setler [38]

Can vary. Carbon is used quite commonly, and extracts metal oxides, works with zinc, iron, tin, lead and copper.

5 0
3 years ago
A solid mixture consists of 47.6g of KNO3 (potassium nitrate) and 8.4g of K2SO4 (potassium sulfate). The mixture is added to 130
IgorLugansk [536]

<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.

<u>Explanation:</u>

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be \frac{7.4}{100}\times 130=9.62g

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

7 0
3 years ago
A 225 g sample of an unknown solid is heated 67C and placed into a calorimeter containing 25.6 g of water at 15.6°c. If the fina
Stella [2.4K]

Answer:

= 1.271 J/g°C

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the solid;

= 225 g × c ×(67 -53) , where c is the specific heat capacity of the metal

= 3150 c joules

Heat absorbed by water;

= 25.6 g × 4.18 J/g°C × (53-15.6)

= 4002.0992  joules

Therefore;

3150 c joules = 4002.0992 joules

c =4002.0992/3150

 <u> = 1.271 J/g°C</u>

6 0
3 years ago
Hydrogen produced from a hydrolysis reaction was collected over water and the following data was compiled.
Shkiper50 [21]

Answer:

  • 0.00358 mol

Explanation:

<u>1) Data:</u>

a) V = 93.90 ml

b) T = 28°C

c) P₁ = 744 mmHg

d) P₂ = 28.25 mmHg

d) n = ?

<u>2) Conversion of units</u>

a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter

b) T = 28°C = 28 + 273.15 K = 301.15 K

c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm

d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm

<u>3) Chemical principles and formulae</u>

a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.

b) Ideal gas equation: pV = nRT

<u>4) Solution:</u>

a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm

b) Moles of hygrogen gas:

pV = nRT ⇒ n = pV / (RT) =

n =  (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =

n = 0.00358 mol (which is rounded to 3 significant figures) ← answer

7 0
3 years ago
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8
mr_godi [17]

Answer:8

in the respiratory system

a.

Explanation:

7 0
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