What is the empirical formula for the compounds? a. A compound with 92.2% carbon

Question

What is the empirical formula for the compounds? a. A compound with 92.2% carbon and hydrogen b. A compound with 65.4% carbon, 29.1 % oxygen and hydrogen c. A compound with 83.6% carbon and hydrogen d. A compound with 87.7% carbon and hydrogen e. A compound with 36.8% oxygen and manganese f. A compound with 7.21% oxygen and rhodium g. A compound with 62.5% calcium and carbon h. A compound with 87.4% nitrogen and hydrogen What is the molecular formula for the previously described compounds with the molar masses? a. 78.12 g middot mol^-1 b. 110.12 g middot mol^-1 c. 86.20 g middot mol^-1 d. 82.16 g middot mol^-1 e. 86.94 g middot mol^-1 f. 221.82 g middot mol^-1 g. 64.10 g middot mol^-1 h. 32.06 g middot mol^-1 What are both the empirical formula and the molecular formula for the compounds described? a. An 18.63 g sample containing 6.204 g of nitrogen, 57.10% carbon, and hydrogen has a molar mass of 168.28 g middot mol^-1. b. A 273 g sample containing 181.85 g of carbon, 22.19% oxygen, and hydrogen has a molar mass of 504.84 g middot mol^-1. c. A 0.9875 g sample containing 0.3591 g of sulfur, 9.162% hydrogen, and carbon has a molar mass of 264.57 g middot mol^-1. d. A 75.86 g sample containing 28.69 g of carbon, 6.361% hydrogen, and chlorine has a molar mass of 317.55 g middot mol^-1. What is the concentration of the solutions? a. 32.5 mg of ammonium nitrate in 6.5 mL of solution b. 1.25 mmol of silver acetate in 12.5 mL of solution

Explanation / Answer

Qn 7. To find out empirical formula of a compound if mass percentages of the constituent elements are given, do the following steps:

a) 92.2% carbon and 7.8 % hydrogen.

No. of moles of hydrogen = 7.8/1.00797 = 7.738325545

3. simplest mole ratio:

For carbon: 7.676296728 / 7.676296728= 1

For hydrogen: 7.738325545 / 7.676296728 = 1.00808

4. Subscript of carbon = 1

Subscript of hydrogen = 1

Thus empirical formula of the given compound is CH.

Similarly answers for the remaining compounds are as follows:

b) C3H3O

c) CH2

d) CH2

e) MnO2

f) Rh2O

g) CaC2

h) NH2

Qn 8. To find out molecular formula of a compund from its empirical formula and molecular mass, use the following steps:

a) Empirical formula : CH

Therefore molecular weight of the compund = C6H6

Similarly molecualr weights of the remaining compounds are given below:

b) C6H6O2

c) C6H12

d) C6H12

e) MnO2

f) Rh2O

g) CaC2

h) N2H4

Qn 9.

a) Amount of compund = 18.63 g

Step 1 : Mass of nitrogen present = 6.204 g

Mass of carbon present = (18.63* 57.10) / 100 = 10.63773 g

Mass of hydrogen present = 1.78827 g

Step 2 : Moles of each components present:

Moles of nitrogen = 6.204 / 14.0067 = 0.442930883

Moles of carbon = 10.63773 / 12.011 = 0.885665639

Moles of hydrogen = 1.78827/ 1.00797 = 1.774130182

Step 3 : simplest mole ratio

For nitrogen: 0.442930883/ 0.442930883 = 1

For carbon: 0.885665639 / 0.442930883 = 1.9995572

For hydrogen: 1.774130182 / 0.442930883 = 4.005433

Step 4: Round to nearest whole number.

Subscript for nitrogen = 1

Subscript for carbon = 2

Subscript for hydrogen = 4

Thus empirical formula of the compound is C2H4N

To find molecular formula:

Step 1: Empirical fomula weight = (12.011*2) + (1.00797*4) + (14.0067*1) = 42.06058

Step 2: Scaling factor = 168.28 / 42.06058 = 4.0008 = 4

Therefore molecular formula of the compiund is C8H16N4

Follow the same steps to find out the empirical as well as molecular formula of the compounds:

b) Empirical formula: C4H8O ; Molecular formula: C21H32O7

c)  Empirical formula: C4H8S ; Molecular formula: C12H24S3

d)  Empirical formula: C2H4Cl ; Molecular formula:C10H20Cl5

Qn 10.

a) Molarity = moles of solute/ volume of solution in litre

Moles of ammonium nitrate = Mass/ molecular weight = (32.5*10-3) / 80.043 = 4.0603 * 10-4

   Molarity of the solution = ( 4.0603 * 10-4) / (6.5* 10-3) = 0.062 M

b) Molarity = (1.25* 10-3) / (12.5 * 10-3) = 0.1 M

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