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Daniel [21]
3 years ago
5

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

is 820 torr. What is the molecular formula for this compound?
Chemistry
1 answer:
ycow [4]3 years ago
8 0

<u>Answer:</u> The molecular formula for the compound is C_6H_6

<u>Explanation:</u>

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = \frac{7.68}{7.68}=1

For Hydrogen = \frac{7.74}{7.68}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

  • <u>Calculating the molar mass of the compound:</u>

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{m}{M}RT

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L  (Conversion factor:  1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = 62.3637\text{ L. torr }mol^{-1}K^{-1}

T = temperature of the gas = 100^oC=(100+273)K=373K

Putting values in above equation, we get:

820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

n=\frac{78.10g/mol}{13g/mol}=6

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 6)}H_{(1\times 6)}=C_6H_6

Hence, the molecular formula for the compound is C_6H_6

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8 0
2 years ago
3.8 liters of sulfur vapor, S8(g), at 921.4°C and 5.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measur
lilavasa [31]

Answer:

116.5 g of SO₂ are formed

Explanation:

The reaction is:

S₈(g) +  8O₂(g)  → 8SO₂ (g)

Let's identify the moles of sulfur vapor, by the Ideal Gases Law

We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K

5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K

(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈

Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide

Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 =  1.82 moles

We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g

3 0
3 years ago
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Answer:

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rodikova [14]
The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

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  Mn(2+) + 2e-  --->  Mn(s)                      Eo = - 1.18 V
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The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

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