1 What is the empirical formula for a compound that has 30.4% N and - - PDF document

1

Empirical and molecular formulas

% composition = part whole 100

A) Calculate the % composition for CuSO4•5H2O MM = 1 % Cu = % S = % O = % H = MM = 1 63.5

( )

MM = 1 63.5

( ) +1(32)

MM = 1 63.5

( ) +1(32) + 4 16 ( )

MM = 1 63.5

( ) +1(32) + 4 16 ( ) + 5 18 ( ) ) = 249.5 g

mol 63.5 249.5 100 100 = 25.25% % S = 32 249.5 100 % S = 32 249.5 100 = 12.83% 249.5 % O = 9 16

( )

249.5 100 249.5 % O = 9 16

( )

249.5 100 = 57.72% % H = 10 1

( )

249.5 100 % H = 10 1

( )

249.5 100 = 4.01% B) How many g O are in 25.0 g of CuSO4•5H2O

g O g total 100 = mass % g O g total = mass fraction g O = = mass fraction g total = 0.5772 0.5772 25.0 g = 14.43 g O Molecular formula Empirical formula

Two different types of chemical formulas smallest whole # ratio of atoms in comp’d Used for ionic comp’d actual # of atoms in molecule Used for covalent comp’d

Covalent compound

C6H12O6 Molecular formula Empirical formula CH2O C6H16N2 C3H8N C12H4Cl4O2 C6H2Cl2O H2O H2O H2O2 HO Determining empirical formula for a comp’d

• 1. Obtain mass of each element
• 2. Determine the # moles of each
• 3. Divide the # moles of each by the smallest # of moles

to convert the smallest to 1: gives mole ratio

• 4. If one or more of the numbers are not integers, multiply

each by the smallest integer that will convert all into whole numbers. 2, 3, 4, 5

• 5. These #s are the subscripts in empirical formula
• 6. To determine molecular formula, compare ratio of

molecular to empirical formula masses

2

What is the empirical formula for a compound that has 30.4% N and 69.6% O?

Assume 100g 30.4 g N and 69.6 g O 30.4 g N

• Step 2: convert to moles

Step 1: calculate mass of each element:

69.6 g O

• 30.4 g N 1 mole N

14 g N

• = 2.17 mole N

69.6 g O 1 mole O 16 g O

• = 4.35 mole O

Step 3: get mole ratio by dividing each number by the smaller of the two

2 O for every 1 N 2.17 NO2 2.17 = 1 mole N 4.35 2.17 = 2 mole O

Find MM of empirical formula unit and compare to molecular

• Find MM of empirical formula

NO2 =

92 46= 2

• Compare the two MM:

Comp’d has 2 x MM of emp. form.:

N2O4

Step 6

46 g/mol For above problem, MM of molecular formula = 92 g/mol What is the empirical formula for a compound that contains 65.2% arsenic and 34.8% oxygen?

Assume 100g 65.2 g As and 34.8 g O

Step 2: convert to moles Step 1: calculate mass

65.2 g As

• 34.8 g O
• 65.2 g As 1 mole As

74.9 g As

• = 0.870 mole As

34.8 g O 1 mole O 16 g O

• = 2.18 mole O

As2O5

Step 3: get mole ratio by dividing by smallest Step 4: Multiply each by small integer to get whole #

2 arsenic for every 5 oxygen

0.87 0.87 87 87 = 1 mole As 2.18 0.87 = 2.5 mole O

X 2

For the previous problem, if MM is 230 g/mol, what is its molecular formula?

2 75 = 150g

As2O5 = molecular formula

Step 6: Find MM of empirical formula unit and compare to molecular

230 230 = 1 1 empirical formula unit in molecular

5 16 = 80g 230 g/mol

3

A compound contains 73.14 % C and 7.37% H. The rest is

• O. A 10.0 g sample was found to contain .06165 moles.

What are the empirical and molecular formulas?

Step 1: mass % to grams: Step 2: Convert mass to moles:

100 g = 73.14 g C 73.14 g C

• 73.14 g C

1 mole C 12.011 g C

• = 6.089 mole C

7.37 g H

• 7.37 g H 1 mole H

1.0079 g H

• = 7.31 mole H

19.49 g O

• 19.49 g O 1 mole O

16.0 g O

• = 1.218 mole O

73.14 g C + 7.37 g H 7.37 g H + mass O Mass O = 19.49 g

Step 3: Find mole ratio and subscripts

6.089 mole C 1.218 mole O

• Empirical formula = C5H6O

7.31 mole H 1.218 mole O

• = 5 mole C
• = 6 mole H

MM = 5(12)+6(1)+1(16) = 82

Step 6: calculate molecular mass

Molecular formula = C10H12O2 10.0 g 0.06165 mol = 162.2 g/mol

A 10.0 g sample was found to contain 0.06165 moles. What are the empirical and molecular formulas?

162.2 82 = ~ 2

7) A nonmetal oxide compound is 50.0% by mass oxygen and has an empirical formula of XO2. Identify X.

XO2 50% by mass O assume 1 mole of compound : let x = MM of element X 2(16) 32 = x = 32g X 2 mol O 1 mol X 2(16) 2(16) + x x = 0.5 16+ 0.5x X = S SO

S SO2

8) A nonmetal oxide compound is 74.1% by mass oxygen annd has an empirical formula of X2O5. Identify X.

X2O5 74.1% by mass O assume 1 mole of compound : 5 mol O 2 mol X let x = MM of element X 5(16) 80 = x = 13.98g X 5(16) 5(16) + 2x = 2x = 0.741 59.28 .28 +1.482x X = N N

N N2O5