Quiz 6 - Percent Composition and Empirical Formulas

The balanced equation tells you how much reactant you need and allows you to predict how much product you'll make

Review:

Why is it necessary to balance chemical equations?

1. Balance equation

Fe2O3 (s) + 2Al (s) → 2Fe (I) + Al2O3 (s)

2. Find mass of Fe2O3

12.0g Fe (1 mol Fe/ 55.845g Fe) (1 mol Fe2O3/ 2 mol Fe) (156.69g Fe2O3/ 1 mol Fe2O3)

= 17.2g Fe2O3

3. Find mass of Al

12.0g Fe (1 mol Fe/ 55.845g Fe) (2 mol Al/ 2 mol Fe) (26.982g Al/ 1 mol Al)

= 5.80g Al

Review:

What masses of Iron (III) oxide (MM = 159.69 g/mol) and aluminum (MM = 26.982 g/mol) must be used to produce 12.0 g of Iron (MM = 55.845 g/mol)? (balance the equation)

Fe2O3 (s) + Al (s) → Fe (I) + Al2O3 (s)

Molecular mass

____________ is the sum of atomic masses in a molecular formula

Percent X = (mass X / mass sample) x 100%.

What equation is used to find percent composition

% C = (2.434g C / 3.789g sample) x 100%

= 64.24% C

% H = 100% - 64.24%

= 35.76% H

Find the percent composition:

an unknown solid of 3.789g decomposes into 2.434g of C and 1.355g of H

Theoretical percent composition:

%O = (15.999g O / 44.013g N2O) x 100% = 36.35%

%N = (2 x 14.007g N / 44.1013g N2O) x 100% = 63.65%

Percents are consistent with N2O

a compound containing N and O was found to be 36.3% O and 63.7% N. Is this consistent with N2O

2(14.007) + 5(15.999) = 108.009g N2O5

%N = (2 x 14.007g N / 108.009g N2O5) x 100% = 25.94% N

% O = (5x 15.999g O / 108.009g N2O5) x 100% = 74.06% O

Find the theoretical percent composition of N2O5

Empirical formula

The _____________ is the Smallest whole number ratio of elements

Molecular formula

The ______________ is the actual chemical formula for a molecular compound

P4O10 - molecular formula

P2O5 - empirical formula

Identify which is the empirical formula and which is the molecular formula

P4O10

P2O5

Ionic compounds

___________ only possess empirical formulas, not molecular

False, can be the same (ex. N2O and N2O5)

True or False:

Empirical and molecular formulas cannot be the same

1. Mass data

2. Percent composition

3. Combustion analysis

What 3 methods can be used to find the empirical formula?

3.44g compound - 1.52g Fe = 1.92g Cl

1. Find moles

1.52g Fe (1 mol Fe / 55.85g Fe) = 0.272 mol Fe

1.92g Cl (1 mol Cl / 35.45g Cl) = 0.0542 mol Cl

2. Find ratio of moles

(0.0542 mol Cl / 0.0272 mol Fe)= 2 Cl/ 1 Fe

3. Write empirical formula

FeCl2

Using mass data find the empirical formula:

3.44g of compound containing Fe and Cl is determined to consist of 1.52g Fe

1. Find moles

2. Find mole ratios

(0.1769 mol Fe / 0.2650 mol O) = 1 Fe/ 1.5 O

2 (1 Fe/ 1.5 O)

3. Write empirical formula

Fe2O3

Using mass data find the empirical formula:

9.88g Fe and 4.24g O

Assume 100g of sample 69.6% → 69.6g Ba

1. Find moles

69.6g Ba (1 mol Ba / 137.33g Ba) = 0.507 mol Ba

6.09g C (1 mol C / 12.011g C) = 0.507 mol C

24.3g O (1 mol O / 15.999g O) = 1.52 mol O

2. Find mole ratios

(0.507 mol C / 0.507 mol Ba) = 1 C/ 1 Ba

(1.52 mol O / 0.507 mol Ba) = 3 O/ 1 Ba

3. Write empirical formula

BaCO3

Using percent composition find the empirical formula:

69.6% Ba

6.09% C

24.3% O

1. Find mass

2.469g CO2 (1 mol CO2 / 44.009g CO2) (1 mol C / 1 mol CO2) (12.011g C/ 2 mol C) = 0.6738g C

1.009g H2O (1 mol H2O / 18.015g H2O) (2 mol H/ 1 mol H2O) (1.0078g H/ 1 mol H) = 0.1129g H

1.683g - 0.6738g C - 0.1229g H = 0.8963g O

2. Find moles

0.638g C (1 mol C/ 12.011g C) = 0.05610 mol C

0.1129g H (1 mol H/ 1.0079g H) = 0.1120 mol H

0.8963g O (1 mol O/ 15.999g O) = 0.05602 mol O

3. Find mole ratios

(0.05610 mol C/ 0.05602 mol O) = 1 C/ 1 O

(0.1120 mol H/ 0.05602 mol O) = 2 H/ 1 O

4. Write empirical formula

CH2O

Using combustion analysis find the empirical formula:

a compound containing C, H and O weighing 1.683g is combusted to give 2.469g CO2 and 1.009g H2O

1. Find Mass

2.829g CO2 (1 mol Co2/ 44.009c CO2) (1 mol C/ 1 mole CO2) (12.011g C. 1 mol C) = 0.7721g C

1.159g H2O (1 mol H2O/ 18.015g H2O) (2 mol H/ 1 mol H2O) (1.0078g H/ 1 mol H) = 0.1300g H

1.005g - 0.7721g C - 0.1300g H = 0.1029g O

2. Find Moles

0.7721g C (1 mol C/ 12.011g C) = 0.06248 mol C

0.1300g H (1 mol H. 1.0079g H) = 0.1290 mol H

0.1029g O (1 mol O/ 15.999g O) = 0.006431 mol O

3. Find Mole ratios

(0.06428 mol C/ 0.006431 mol O) = 10 C/ 1 O

(0.1290 mol H/ 0.006431 mol O) = 20 H/ 1 O

4. Write empirical formula

C10H20O

Using combustion analysis find the empirical formula:

a compound containing C, H and O

1.005g sample

2.829g CO2

1.159g H2O

molecular formula = Molar mass / empirical mass

What equation is used to find the molecular formula?

empirical mass = element 1 MM + element 2 MM

What equation is used to find the empirical mass?

1. empirical mass = C MM + H MM

12.011 g/mol C + 1.008 g/mol H = 13.019 g/mol

2. molecular formula = Molar mass / empirical mass

(104.15 g/mol) / (13019 g/mol) = 7.99 = 8

8 (CH) = C8H8

Find the molecular formula:

empirical formula → CH

MM = 104.15 g/mol

empirical mass → 156.3 g/mol (312.5 g/mol) / (156.3 g/mol) = 2

2 (C10H20O) = C20H40O2

Find the molecular formula:

empirical formula → C10H20O

MM → 312.5 g/mol

a) Empirical formula:

1. Find mass

220.0 mg CO2 (10^-3g CO2/ 1mg CO2) (1mol CO2/ 44.011g CO2) (1 mol C/ 1 mol CO2) (12.011g C/ 1 mol C)

= 0.06004g C

45.0 mg H2O (10^-3g H2O/ 1mg H2O) (1 mol H2O/ 18.015g H2O) (2 mol H/ 1 mol H2O) (1.008g H/ 1 mol H)

= 0.00504g H

81.0 mg (10^-3g/ 1mg)

0.0810g - 0.06004g - 0.00504g = 0.0159g O

2. Find moles

0.06004g C (1 mol C/ 12.011g C) = 0.005000 mol C

0.00504g H (1 mol H/ 1.008g H) = 0.00500 mol H

0.0159g O (1 mol O/ 15.999g O) = 0.000994 mol O

3. mole ratios

(0.005000 mol C/ 0.000994 mol O) = 5/1

((0.00500 mol C/ 0.000994 mol O) = 5/1

C5H5O → empirical formula

b) Molecular formula:

(5 x 12.011) + (5 x 1.008) + 15.999 = 81.094 g/mol → empirical mass

(243 g/mol)/ (81.09 g/mol) = 3

3 (C5H5O) = C15H15O3 → molecular formula

Find the empirical and molecular formula:

The combustion of 81.0 mg of a compound (MM = 243 g/mol) extracted from a plant and known to contain C, H, and O produces 220.0 mg CO2 and 45.0 mg H2O

Limiting reactant

The _____________ reactant is used up first in the reaction and tells us how much product can be made

Excess reactant

The _______________ reactant is what is left over after the reaction (more than needed)

1. Determine limiting reactant (LR)

a) 2.00g KOH → ? g CuCl2

2.00g KOH (1 mol KOH/ 56.qqg KOH) (1 mol CuCl2/2 mol KOH) (134.45g CuCl2/ 1 mol CuCl2)

= 2.40g CuCl2 → required

We are given 3.00g of CuCl2 → have more than needed

- CuCl2 is in excess

- KOH is the limiting reactant

b) 3.00g CuCl2 → ? g KOH

= 2.50g KOH → required

Only 2.00g KOH is given → not enough

- CuCl2 is in excess

- KOH is the limiting reactant

2. Calculate the amount of product using the amount of given limiting reactant (LR)

2.00g KOH (1 mol KOH/ 56.11g KOH) (1 mol Cu(OH)2/ 2 mol KOH) (96.56g Cu(OH)2/ 1 mol Cu(OH)2)

= 1.72g Cu(OH)2

How many grams of Cu(OH)2 (MM = 96.56 g/mol) can be made from 2.00g KOH (MM = 56.11 g/mol) and 3.00g CuCl2 (MM = 134.45 g/mol)?

CuCl2 (aq) + 2KOH (aq) → Cu(OH)2 (s) + 2KCl (aq)

1. Determine the LR

1142g C6H5Cl (1 mol C6H5Cl/ 112.56g C6H5Cl) (1 mol C2HOCl3/ 1 mol C6H5Cl) (147.39g C2HOCl3/ 1 mol C2HOCl3)

= 747.7g C2HOCl3 → required

485g C2HOCl3 is given → not enough

- C2HOCl3 is the limiting reactant

- C6H5Cl is in excess

2. Calculate the amount of product using the amount of LR given

485g C2HOCl3 (1 mol C2HOCl3/ 147.39g C2HOCl3) (1 mol DDT/ 1 mol C2HOCl3) (354.49g DDT/ 1 mol DDT)

= 1166g DDT → 1170g DDT → theoretical yield

If 1142g of C6H5Cl (MM = 112.56 g/mol) were reacted with 485g of C2HOCl3 (MM = 147.39 g/mol) how many grams of DDT (MM = 354.49 g/mol) should be made?

- actual yield

- theoretical yield

The ________________ is the amount of product actually obtained in the reaction, while the ______________ is the maximum amount of product that can be obtained from a given amount of reactant

Limiting reactant

You need to know the _________________ to find the theoretical yield. (You cannot solve for percent yield without knowing the ______________)

Percent yield = (actual yield / theoretical yield) x 100%

What equation is used to find percent yeild?

Percent yield = (954g/ 1170g) x 100% = 81.5%

Find percent yeild:

954g DDT was made

1170g DDT → theoretical yield

1. Find Limiting Reactant and excess reactant

68.5kg CO (10^3g CO/ 1 kg CO) (1 mol CO/ 28.01g CO) (2 mol H2/ 1 mol CO) (2.016g H2/ 1 mol H2)

= 9860g H2

→ 8600g H2 is given, there is not enough

- H2 is the limiting reactant

- CO is the excess reactant

2. Find the theoretical yield (product) using the given LR

8600g H2 (1 mol H2/ 2.016g H2) (1 mol CH3OH/ 2 mol H2) (32.03g CH3OH/ 1 mol CH3OH)

= 68300g CH3OH

3. Find percent yield

3.57 x 10^4 mL (0.792g/ 1 mL) = 28300g → actual yield

(28300g/ 68300g) x 100% = 41.4%

Find the Limiting reactant, excess reactant, theoretical yield and percent yield:

The reaction of 68.5 kg of CO and 8.60 kg of H2 produces 3.57 x 104 mL of methanol (CH3OH). (Density of methanol = 0.792 g/mL)

CO (g) + 2H2 (g) → CH3OH (l)