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Unit 6

Lesson 25- Molecular Mass and Moles

Lesson 26 – Percentage Composition

Lesson 27 – Empirical Formula

Lesson 28 – Molecular Formulas

Lesson 25 - Molecular Mass and Moles

Learning Targets:

· The student will determine the molecular mass of a formula.

· The student will convert between mass and moles for a formula.

Chemistry: Matter and Change Text References

Chapter 11; pages 309-327

I. Molecular Mass

A. The of a formula is the mass of that formula expressed in grams per mole.

B. This is calculated by the masses of all of the elements contained in the formula for a compound.

C. Example: H2O contains 2 hydrogen atoms and 1 oxygen atom. When we look on the periodic table,

i. Oxygen = 16.00 g/mol

ii. Hydrogen = 1.01 g/mol

D. Therefore, 2 x 1.01 + 1 x 16.00 = 18.02 g/mol for water, or H2O.

E. Examples:

CO2H2O2

F. If a formula contains parenthesis with subscripts following, the subscript should be through the parenthesis to calculate the number of atoms of each element

G. Examples: Mg3(PO4)2

II. Conversion between mass and moles

A. The conversion between mass and moles of a substance is accomplished using .

B. Use the molecular mass of the compound as the conversion factor – example: 18.02 g H2O / 1 mol H2O. Flip it upside down if needed to cancel units.

C. Examples:

10g H2O84 mols O2

III. Conversion between moles and particles

A. The conversion between moles and number of particles ( atoms, molecules, or formula units) is also accomplished with

B. Avagadro’s Principle states that there are particles in one mole. This is then becomes a conversion factor with Avagadro’s number and 1 mole as the two terms.

C. The direction in which you are converting determines which goes on top.

D. Examples:

10 mol H2O

1.23 x 1062 molecules of Rd1.23 x 1062 molecules of O2

IV. Conversion between mass and number of particles

A. There is no direct conversion between mass and number of particles.

B. This must be done through a process.

C. First convert mass to moles as described in II above.

D. Then convert moles to particles as described in III above.

Formula Mass Problems

Level 1

1. Find the molecular mass of each of the following compounds.

a. KCl

b. NaF

c. HI

d. LiBr

e. RbBr

f. BaCl2

g. CaI2

h. Na2S

i. MgS

j. AlP

2. Find the mass of each of the following, expressed in grams.

a. 1.00 mol of CaO

b. 1.00 mol of BeSe

c. 1.00 mol of KF

d. 1.00 mol of SrO

e. 2.00 mol of MgI2

f. 2.00 mol of Li3P

g. 5.00 mol of CaCl2

h. 0.50 mol of FeBr2

i. 0.20 mol of Cu2O

j. 0.40 mol of Hg2Cl2

3. Find the amount of moles in each of the following masses.

a. 18.02 g of H2O

b. 80.92 g of HBr

c. 17.04 g of NH3

d. 190.42 g of MgCl2

e. 36.75 g of K2S

f. 25.38 g of SnI2

g. 11.98g of FeO

h. 15.0 g of KBr

i. 25.0 g of SrS

j. 50.0 g of AlF3

4. Find the amount of formula units (particles) of each of the masses listed in #3.

Level 2

1. Find the molecular mass of each of the following compounds.

a. HNO3

b. Fe2O3

c. H3PO4

d. K2SO4

e. Be5As2

f. NH4NO3

g. RbSO3

h. Li2CO3

i. Mg(OH)2

j. Al2(SO4)3

2. Find the mass of each of the following, expressed in grams.

a. 1.00 mol of HC2H3O2

b. 2.50 mol of K2CrO4

c. 0.50 mol of Ca(ClO3)2

d. 0.25 mol of Ba(NO3)2

e. 0.375 mol of Na2Cr2O7

f. 0.25 mol of NaC2H3O2

g. 0.152 mol of H3PO4

h. 0.0582 mol of Li2SO4

i. 0.418 mol of Fe(NO3)3

j. 1.872 mol of Cu(C2H3O2)2

3. Find the amount of moles in each of the following masses.

a. 100.00 g of CaCO3

b. 100.00 g of Ni(NO3)2

c. 50.00 g of C6H12O6

d. 25.00 g of K3PO4

e. 15.57 g of Bi(OH)3

f. 3.50 g of AsCl3

g. 0.572 g of Ca3P2

h. 1.750 g of Ca(C2H3O2)2

i. 4.904 g of Al(NO3)3

j. 27.85 g of Fe3(PO4)

5. Find the amount of formula units (particles) of each of the masses listed in #3.

Lesson 26 – Percentage Composition

Learning Targets:

· The student will determine percentage composition from formula data.

· The student will determine percentage composition of a hydrated salt.


Chapter 11; pages 328-329, 338-341

I. Percentage Composition

a. If you need to know how much of a particular element you can get from a compound, you can determine percentage composition

b. Remember that molar mass is ____________________ represented by the compound, you can determine percentage composition

c. Therefore, the percentage of each element in the compound is the molar mass of the element divided by the molar mass of the compound, times 100%.

d. Example:

Copper (I) sulfide is found in nature as the mineral chalcocite, a copper ore.

What is the percent composition of pure chalcocite?

1. Determine the compound formula:

2. Determine the molar masses of the elements:

3. Determine the molar mass for Cu2S:

4. Calculate the percent composition for each element in the compound:

Percentage Cu =

Remember that a percent is:

Percentage S =

Check your work!!! – the percentages should add up to no more than 101% and no less than 99% (due to rounding)

II. Percentage composition tells how much water is in a hydrate

A. Some salts have the ability to bind water molecules within their lattice structure

B. These salts are known as ______________________

C. The anhydrous forms, or forms without water, have various uses:

i.

ii.

D. One salt that can form a hydrate is CuSO4, or copper (II) sulfate

i. Copper sulfate binds 5 water molecules into its structure, forming copper sulfate

pentahydrate.

ii. The formula is written

iii. It indicates that 5 water molecules are included for each formula unit of the main compound.

E. Another example is sodium carbonate decahydrate, or Na2CO3 . 10H2O.

F. Note that the hydrate contains less of the salt than the anhydrous form, because some of the mass of the compound is water in the hydrate, while all of the mass of the compound is the salt in the anhydrous form.

G. How to determine the percentage composition of hydrates –

Example:

What percentage of hydrated sodium carbonate, Na2CO3 . 10H2O, is Na2CO3?

1. Determine the molar mass of Na2CO3

2. Determine the molar mass of 10 moles of H2O

3. Determine the molar mass for Na2CO3 . 10H2O

4. Determine the percentage of Na2CO3 in the hydrate

III. Percent Composition

A. If given the chemical formula for the compound:

1. First calculate the molecular mass of the entire compound

Example: H20

H 2 x 1.0079 = 2.0158

O 1 x 15.999 = 15.999

18.0148 g/n 18.015 g/n with proper sig fig

2. Next divide the mass of each element total by the molecular mass to get the percent composition of each of the elements. Multiply by 100 to get a percent.

H = 2.0158/ 18.015 = .111897 x 100 = 11.190 %

O = 15.999/ 18.015 = .888103 x 100 = 88.810 %

B. If given experimental data in grams follow the same procedure without calculating the molecular weight. Use the individual masses of each element as the element totals, and the total mass of all the elements as the molecular mass.

Example: What is the percent composition of an ore which contains 1.86 grams of aluminum

and 1.65 grams of oxygen?

Total = 1.86 + 1.65 = 3.51grams

Al 1.86 / 3.51 = .5299 x 100 = 53%

O 1.65 / 3.51 = .4700 x 100 = 47%

Percentage Composition Problems

Level 1

Find the percentage composition of each compound listed below. In the first ten problems, the correct formula is given. In the next four problems, laboratory data for the compound are presented.

1. FeO

2. MgCl2

3. CH4

4. CS2

5. NO2

6. HgO

7. SnI4

8. Cu2O

9. NH3

10. CH2O

11. What is the percentage composition of an ore which contains 1.86 g of aluminum and 1.65 g of oxygen?

12. A compound of silver contains 4.35 g of that metal combined with 0.65 g of sulfur. What is the percentage composition of this compound?

13. A certain sample of a gas weighing 22.8 g contains 6.22 grams of carbon and an unweighed quantity of oxygen. What is the percentage composition of this gas?

14. A compound consisting of aluminum and chlorine weighs 17.82 g. The aluminum in the compound weighs 3.60 g. What is the weight of chlorine in the compound? What is the percentage composition of the compound?

Level 2

Find the percentage composition of each compound listed below. In the first eight problems, the correct formula or name is given. In the next six problems, laboratory data for the compound are presented.

1. KNO2

2. NH4Cl

3. SrCl2

4. KMnO4

5. sulfuric acid

6. potassium phosphate

7. ammonium bromide

8. barium hydroxide

9. Analysis of a compound shows that it consists of 43.40 g of copper and 10.95 g of sulfur. What is the percentage composition of this compound?

10. A sample of benzene is analyzed and found to consist of 13.74 grams of carbon and 1.15 grams of hydrogen. What is the percentage composition of benzene?

11. Analysis of an unknown compound shows that it consists of 21.8 g of oxygen, 4.09 g of aluminum, and 6.36 grams of nitrogen. What is the percentage composition of this compound?

12. A compound consisting of carbon, hydrogen, and oxygen weighs 40.85 g. Analysis shows that the compound contains 10.90 g of carbon and 0.90 g of hydrogen. What is the percentage composition of this compound?

13. An organic compound consisting of carbon, hydrogen, and oxygen only weighs 13.669 g. Analysis shows that the compound contains 0.547 g and 8.707 g of the last two of these elements, respectively. What is the percentage composition of this compound?

14. Analysis of an ore of calcium shows it to contain 13.61 g of calcium and 21.77 g of oxygen in a sample weighing 35.38 g. What is the percentage composition of this compound?

Level 3

Find the percentage composition of each compound listed below. In the first eight problems, the correct name is given. In the next six problems, laboratory data for the compound are presented.

1. zinc carbonate

2. ammonium sulfate

3. iron (III) oxide

4. calcium phosphate

5. ammonium dichromate

6. aluminum nitrate

7. calcium acetate

8. ammonium carbonate

9. A sample of water weighing 89.6 g is found to contain 9.98 g of hydrogen. How much oxygen is present in the water? What is the percentage composition of this compound?

10. An organic compound is found to contain 10.18 g of carbon, 1.47 g of hydrogen, and 0.93 g of oxygen. What is the percentage composition of this compound?

11. A sample of organic material weighing 0.005587 g is analyzed and found to contain 2.51 x 10-4 g of hydrogen and 1.98 x 10-3 g of oxygen. What is the percentage composition of this compound?

12. Analysis of 0.005644 g of a compound shows that it contains 4.177 x 10-3 g of carbon and 9.76 x 10-4 g of nitrogen. The hydrogen in the compound cannot be determined directly. What is the percentage composition of this compound?

13. Laboratory analysis of a sample of paraldol shows that it contains 4.67 g of carbon, 0.77 g of hydrogen, and 3.12 g of oxygen. What is the percentage composition of paraldol?

14. An organic compound separated into its elements produces 8.2 g of carbon and 1.44 g of hydrogen from a sample weighing 32.8 g. What is the weight of oxygen in this compound? What is the percentage composition of this compound?

Lesson 27 – Empirical Formula

Learning Target:

· The student will determine an empirical, or simplest, formula from percentage composition data.



I. Formulas can be determined from composition data

A. When a new substance is discovered, the discoverers don’t usually know the formula

B. They do know the percentage composition

C. This can be used to determine the formula, by converting everything from percentages to moles.

D. These moles can then be divided by the smallest number that is determined, and this will give small, whole numbers which can be used to determine ratios.

E. Example

1) As part of a science fair project, Antonio is analyzing the contents of fresh alkaline batteries. He has

determined that one ingredient is a black powdery compound of 63% manganese and 37% oxygen. What is the compound’s formula?

a) List the percentages

Mn = 63%

O = 37%

b) Determine the number of moles of each element present

– do this by assuming you have 100 g of the compound in question.

c) Select the smallest number of moles, and divide all your number of moles by that one:

1.0 mol Mn and 2.1 (which is close to 2) mol O translates into:

II.Recap:

1. % to grams (assume 100g)

2. grams to moles

3. divide by the smallest

III. Special cases

A. Sometimes when you get to the last step, you don’t get numbers close to whole numbers. The following chart should help you with this:

Range

Action

0.1-0.2

Round down

0.3

Multiply everything by 3

0.4-0.6


.7


0.8-0.9

Round up

B. When completed, you should have moles close enough to whole numbers to make an empirical

formula.

IV. Examples:

Empirical Formula Calculation

The empirical formula is the formula with the lowest whole number ration between the elements. Example: the empirical formula of glucose which is C6H12O6 would be CH2O.

A. To calculate the empirical formula from the percent composition:

1. Change the % signs to grams.

2. Convert grams to moles

3. Divide all the numbers of moles by the lowest number of moles.

4. This gives you the ratio of the number of elements in the empirical formula.

Example: 11% Hydrogen and 89% Oxygen

1. 11g H and 89 g O

2. 11g H ( 1 mole/ 1.0079g) = 10.91378 moles

89 g O( 1 mole/ 15.999g) = 5.562848 moles

3. Oxygen has the smaller number of moles

H 10.91378/ 5.562848 = 1. 96 = about 2

O 5.562848/5.562848 = 1

4. Ratio is 2 H’s for every 1 O so the empirical formula is H2O

B. To calculate the empirical formula from experimental data:

1. convert the grams given into moles for each element

2. follow same procedure as above.

Empirical Formula Problems

Objectives:

· The student will convert percentage composition data into an empirical formula.

· The student will use supplied data to calculate a missing piece of data to complete the problem.

Level 1

Determine the simplest formula for each compound listed below. In the first eight problems, the percentage composition of the compound is given. In the last two, laboratory data for the compound are presented.

1. 80.0% carbon, 20.0 % hydrogen

2. 71.5% calcium, 28.5% oxygen

3. 82.2% nitrogen, 17.8% hydrogen

4. 85.7% carbon, 14.3% hydrogen

5. 6.6% aluminum, 93.4% iodine


7. 23.5% potassium, 76.5% iodine

8. 50.0 % sulfur, 50.0% oxygen

9. An oxide of arsenic contains 3.26 g of arsenic and 1.04 g of oxygen. What is the empirical formula for this oxide?

10. A sample of sodium oxide weighing 12.57 g contains 9.34 g of sodium. What is the empirical formula for this compound?

Level 2

Determine the simplest formula for each compound listed below. In the first six problems, the percentage composition of the compound is shown. In the last four, laboratory data for the compound are given.

1. 63.5% silver, 8.2% nitrogen, 28.2% oxygen

2. 14.3% nitrogen, 4.1% hydrogen, 81.6% bromine

3. 24.7% potassium, 34.7% manganese, 40.5% oxygen

4. 56.6% potassium, 8.68% carbon, 34.7% oxygen

5. 9.92% carbon, 58.7% chlorine, 31.4% fluorine

6. 37.8% carbon, 6.4% hydrogen, 55.8% chlorine

7. 35.0% nitrogen, 5.0% hydrogen, 60.0% oxygen

8. 27.4% sodium, 1.2% hydrogen, 14.3% carbon, 57.1% oxygen

9. Analysis of a sample of a sulfur acid shows it to contain 0.17 g of hydrogen, 2.82 g of sulfur, and 5.67 g of oxygen. What is the simplest formula for this compound?

10. Analysis of a salt results in the following composition: 3.47 g of sodium, 2.12 g of nitrogen, and 7.27 g of oxygen. What is the empirical formula for this salt?

11. A barium salt is found to contain 21.93 g of barium, 5.12 g of sulfur, and 10.24 g of oxygen. What is the simplest formula of this compound?

12. An ore containing zinc, carbon, and oxygen, and weighing 485.35 g is analyzed and found to contain 46.59 g of carbon and 186.37 g of oxygen. What is the simplest formula for this compound?

Level 3

Determine the simplest formula for each compound listed below. In the first six problems, the percentage composition of the compound is given. In the last four, laboratory data for the compound are given.

1. 26.6% potassium, 35.4% chromium, 38.0% oxygen

2. 74.0% carbon, 8.7% hydrogen, 17.3% nitrogen

3. 69.8% iron, 30.2% oxygen


5. 50.8% zinc, 16.0% phosphorus, 33.2% oxygen

6. 21.2% nitrogen, 6.1% hydrogen, 24.3% sulfur, 48.4% oxygen

7. Chemical analysis of a 10.000 g sample of oil of wintergreen shows that it consists of 6.32 g of carbon, 0.53 g of hydrogen, and 3.16 g of oxygen. What is the simplest formula for oil of wintergreen?

8. An acid is analyzed in the laboratory and the following results are obtained: 3.1% hydrogen, 31.6% phosphorus, 65.3% oxygen. What is the simplest formula for this acid?

9. Examination of 3.2 x 10-2 g of an unknown white powder shows that the powder consists of an unknown amount of nitrogen, 2.6 x 10-3 g of hydrogen, 6.7 x 10-3 of phosphorus, and 1.37 x 10-2 g of oxygen. What is the simplest formula for this compound?

10. A rock sample weighing 5.88 x 104 g is known to contain calcium, phosphorus, and oxygen. The amount of the first two elements in this rock is found to be 2.28 x 104 g and 1.18 x 104 g respectively. What is the formula for the compound in this rock sample?

Lesson 28 – Molecular Formulas

Objectives:

· The student will calculate molecular formulas from an empirical formula and a molecular mass.

· The student will calculate empirical and molecular formulas from percentage composition data and a molecular mass.



I. Empirical Formulas

A. Up to this point, we have represented formulas in two different ways:

1. _______________________

2. _______________________

B. Now, we will look at the difference between empirical formulas and Molecular formulas

II. Empirical formulas are determined experimentally

A. Many times when scientists discover a new compound, they determine the empirical formula from

_______________________.

B. They can also use other methods to determine the molecular mass of the compound present. This can be

used to determine a molecular formula.

III. Molecular formulas

A. An empirical formula just tells you how much of each element is present, but in lowest number ratios.

B. ___________________________ – gives type and actual number of elements in a compound.

C. Many different compounds can have the same empirical formula, but they may not have the same molecular formula.

D. Also, many times the empirical formula for a compound is the same as the molecular formula

– sometimes, it is not.

E. Examples – all these have CH2O as their empirical formula

1.

2.

3.

IV. Molecular formulas are determined from molar mass

A. Molecular formulas are __________________ of the empirical formulas

B. X (empirical formula) = molecular formula

C. X will be a whole number.

D. To determine X, divide the molecular formula mass by the empirical formula mass

Examples:

Molecular Formula Calculations:

In calculations of molecular formula, you will be given percent compositions and the molecular weight of the entire compound.

1. Calculate the empirical formula as shown above.

2. Calculate the empirical mass of the empirical formula.

3. Divide the given molecular mass by the empirical mass.

4. The answer should be a whole number, multiply this through the subscripts of the empirical formula to get the molecular formula.

Example: Problem 1 level 1

26.7% C = 26.7g C (1 mole/ 12.011)= 2.16468moles C

2.2% H= 2.2g H(1 mole/ 1.0079)= 2.182756moles H

71.1%O=71.1g O(1 mole/15.999)= 4.444 moles O

2.16468 is the smallest so divide all by this

C 2.16468/2.16468= 1

H 2.182756/2.16468 = 1.008 = about 1

O 4.444/ 2.16468 = 2.053 = about 2

1. Use these numbers as the subscripts, giving you CHO2

2. Now take this formula to calculate the empirical mass

C 1 x 12.011=12.011

H 1 x 1.0079= 1.0079

O 2 x 15.999=31.998

45.169 empirical mass

3. now divide the molecular mass you were given in the problem, by the empirical mass that you calculated:

90/45.169 = 2

4. take this number and multiply all the subscripts in the empirical formula by it to get the molecular formula

CHO2 becomes C2H2O4

Example: A refrigerant, Freon-12, is composed of 9.93% Carbon, 58.64% Chlorine, and 31.43% Fluorine. What is its molecular formula is it has a molar mass of 120.91g/n?

Lesson 28 - Molecular Formula

Objectives:

· The student will use percentage composition data to calculate an empirical formula.

· The student will use molecular mass data to determine a molecular formula from their determined empirical formula.

Level 1

1. A compound (CH4N) has a molar mass of 60.10g/mol. Find the molecular formula

2. A compound has an empirical formula of PbCl4 and a molecular formula mass of 349.0 g/n. What is its true formula?

3. The percentage composition of ethane gas is 80.0% carbon and 20.0% hydrogen. The molecular weight for ethane is 30μ. What is the correct formula for this compound?

4. A compound was found to contain 49.98 g carbon and 10.47 g hydrogen. The molar mass of the compound is 58.12 g/mol. Determine its molecular formula.

5. A compound has the following percentage composition: 26.7% carbon, 2.2% hydrogen, 71.1% oxygen. The molecular weight of this compound is 90g/n. What is the compound’s true formula?

6. What is the molecular formula if a sample contains 10.52 g Ni, 4.38 g C, 5.10 g N, and a molar mass of 332.2 g/n?

7. A certain compound was analyzed and found to have the following composition: 54.6% carbon, 9.0% hydrogen, and the rest is oxygen. The true molecular weight for the compound is 176g/n. What is the molecular formula of the compound?

8. An unknown compound is analyzed and found to consist of 49.0% carbon, 2.7% hydrogen, and 48.2% chlorine. Boiling point data suggest that the molecular weight of this compound is around 150g/n. What molecular formula would you predict for this compound?

9. Monosodium glutamate (MSG) is sometimes added to food to enhance flavor. Analysis determined this compound to be 35.5% C, 4.77% H, 8.29% N, 13.6 % Na, 37.9% O, and a molecular mass of 169.111 g/n. What is the molecular formula for MSG?

10. Determine the molecular formula for caffeine if it has a composition of 49.48% carbon, 5.19% hydrogen, 16.48% oxygen, 28.85% nitrogen, and a molar mass of 194.14g/n

11. An unknown sample is found to have 14.4 g C, 1.6 g H, 11.2 g N, and 12.8 g O. It is also found to have a molecular mass of 401 g/n. What is its true formula?

Level 2

Notes: a) Useful conversion factor: At STP conditions, 1 mole of gas occupies 22.4 L.

b) Calculating moles present from boiling point elevation requires the following formula:

(Change in BP)

Boiling Point ------------------ X (kg of solvent) = moles present.

Elevation (0.5121)

(Change in FP)

Freezing Point------------------ X (kg of solvent) = moles present

Depression (-1.86)

Once you know moles present and grams present, molecular mass determination is simple. Water normally freezes at 0.0oC and boils at 100.0oC.

1. Chemical analysis of a gaseous compound shows its composition to be 36.4% carbon, 57.5% fluorine, and 6.1% hydrogen. A sample of 1.00 L of this gas weighs 3.96 g. What molecular formula do these data suggest for this compound?

2. A gaseous compound consists of 47.3% sulfur and 52.7% chlorine. 500. mL of this gas weighs 3.00 g. From these data, calculate the molecular formula for this gas.

3. One of the oxides of carbon has a composition of 42.8% carbon and 57.2% oxygen. 250. mL of this gas weighs 0.313 g. What is the compound’s true molecular formula?

4. Analysis of an organic compound indicates that it has a percentage composition as follows: 40.7% carbon, 5.0% hydrogen, 54.3% oxygen. When this compound is vaporized, 35.0 mL of the vapor weighs 0.184 g. What molecular formula do you predict for this compound?

5. An organic compound with the following composition is dissolved in water: 51.5% carbon, 8.7% hydrogen, 39.8% nitrogen. When 35.00 g of this compound is dissolved in 250 g of water, the boiling point of the resulting solution is 100.52o C. What molecular formula does this suggest for the compound?

6. A sample of an organic compound weighing 4.585 g is analyzed and found to consist of 3.056 g of carbon, 0.257 g of hydrogen, 0.677 g of oxygen, and 0.595 g of nitrogen. When 86.4 g of this compound is dissolved in 400 g of water, the resulting freezing point is –3.72oC. What molecular formula does this data suggest for the compound?

7. A sample of an unknown compound weighing 3.58 g is analyzed and found to consist of 3.18 g of carbon and 0.40 g of hydrogen. When 4.50 g of this compound is dissolved in 100.0 g of water, the freezing point of the resulting solution is –0.620oC. What molecular formula do these data suggest for this compound?

8. A monosaccharide of unknown composition is analyzed and found to consist of 40.2% carbon, 53.3% oxygen, and 6.6% hydrogen. A solution formed by dissolving 6.67 g of this compound in 150.0 g of water has a freezing point of –0.45oC. What formula do you predict for this compound?

9. Analysis of a gaseous organic compound shows it to consist of 81.9% carbon and 18.2% hydrogen. A sample of 65.0 mL of the gas is found to weigh 0.296 g at STP. What is the correct molecular formula for this compound?

10. An unidentified organic compound has a composition of 40.0% carbon, 6.7% hydrogen, and 53.2% oxygen. When 28.6 g of this compound is dissolved in 245 g of water, the solution has a boiling point of 100.33oC. What molecular formula corresponds to these data?

Unit 6 Exam Study Guide

Percent Composition, Empirical Formula, Molecular Formula

This exam covers the topics related to Lecture lessons 24-27

The problems are very similar to those assigned in class and homework assignments.

Review your lecture notes and rework some of your homework problems to prepare.

Test will consist of:

6 percent composition calculations: 4 directly from a formula, 2 from experimental data.

5 empirical formula calculations.

3 molecular formula calculations.

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