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3 years ago

1) Chemical analysis of a gaseous compound show its composition to be 36.4% carbon, 57.5% fluorine, and 6.1%

Chemistry

1 answer:

Serga [27]3 years ago

5 0

Answer:

1. Empirical formula = CFH₂

Molecular formula = C₉F₉H₁₈

2. Empirical formula is CH₂Cl

Molecular formula = C₂H₄Cl₂

3. Empirical formula = CFH₂

Molecular formula = C₉F₉H₁₈

Explanation:

1) Given data

Percentage of carbon= 36.4%

Percentage of fluorine = 57.5%

Percentage of hydrogen = 6.1%

mass = 296 g

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of C = 36.4/12 = 3.03

Number of gram atoms of F = 57.5/19 = 3.03

Number of gram atoms of H = 6.1/ 1 = 6.1

Atomic ratio:

C : F : H

3.03/3.03 : 3.03/3.03 : 6.1/3.03

1 : 1 : 2

C : F : H = 1 : 1 : 2

Empirical formula is CFH₂

Now we calculate Molecular formula

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 296 / 32

n = 9.25 we take it as 9

Molecular formula = n (empirical formula)

Molecular formula = 9 ( CFH₂)

Molecular formula = C₉F₉H₁₈

Check the answer by calculating the molecular mass, it should be almost 296 g.

Molecular formula = C₉F₉H₁₈

molecular mass = 12 (9) + 18.998 (9) + 1 (18)

molecular mass = 108 + 170.98 + 18

molecular mass = 296.98 g/mol

2) Given data

Percentage of carbon= 24.3%

Percentage of hydrogen = 4.1%

Percentage of chlorine = 71.6%

molecular mass = 99.8 g/mol

Molecular formula = ?

Solution:

Number of gram atoms of C = 24.3/12 = 2

Number of gram atoms of H = 4.1/1 = 4.1

Number of gram atoms of Cl = 71.6/35.5 = 2

Atomic ratio:

C : H : Cl

2/2 : 4/2 : 2/2

1 : 2 : 1

C : H : Cl = 1 : 2 : 1

Empirical formula is CH₂Cl

Now we calculate Molecular formula

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 99.8 / 49.5

n = 2.02 we take it as 2

Molecular formula = n (empirical formula)

Molecular formula = 2 ( CH₂Cl)

Molecular formula = C₂H₄Cl₂

Check the answer by calculating the molecular mass, it should be almost 99.8 g.

Molecular formula = C₂H₄Cl₂

molecular mass = 12 (2) + 1 (4) + 2 (35.5)

molecular mass = 24 + 4 + 71

molecular mass = 99 g/mol

3) Given data

Percentage of carbon= 36.4%

Percentage of fluorine = 57.5%

Percentage of hydrogen = 6.1%

mass = 296 g

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of C = 36.4/12 = 3.03

Number of gram atoms of F = 57.5/19 = 3.03

Number of gram atoms of H = 6.1/ 1 = 6.1

Atomic ratio:

C : F : H

3.03/3.03 : 3.03/3.03 : 6.1/3.03

1 : 1 : 2

C : F : H = 1 : 1 : 2

Empirical formula is CFH₂

Now we calculate Molecular formula

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 296 / 32

n = 9.25 we take it as 9

Molecular formula = n (empirical formula)

Molecular formula = 9 ( CFH₂)

Molecular formula = C₉F₉H₁₈

Check the answer by calculating the molecular mass, it should be almost 296 g.

Molecular formula = C₉F₉H₁₈

molecular mass = 12 (9) + 18.998 (9) + 1 (18)

molecular mass = 108 + 170.98 + 18

molecular mass = 296.98 g/mol

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

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3 years ago

An atom has a mass number of 24 and 13 neutrons. what is the atomic number of this atom?

Elodia [21]

do ityou times 24×13=312.

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3 years ago

How many grams of magnesium acetate are in 8.95x10^23 formula units?

Natasha_Volkova [10]

Answer:

211.63 g.

Explanation:

Particles could refer to atoms, molecules, formula units.

Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole → 6.022 x 10²³ molecules.

??? mole → 8.95 x 10²³ molecules.

The no. of moles of magnesium acetate = (8.95 x 10²³ molecules) (1.0 mole) / (6.022 x 10²³ molecules) = 1.486 mol.

∴ The grams of magnesium acetate are in 8.95 x 10²³ formula units = n x molar mass = (1.486 mol)(142.394 g/mol) = 211.63 g.

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3 years ago