Vanessa Prasad-PermaulValencia College

CHM 1045

Atomic and Molecular Mass

The molecular mass is the sum of the masses of the atoms making up the molecule (units amu) (molecular compounds only!)

calculate the molecular mass:

H2O

C2H4O2

Formula Mass

Formula Mass is the sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic (units amu)

calculate the formula mass:

NaCl

MgCO3

Exercise 3.1 Calculate the formula masses of the

following compounds, using the table of atomic masses. Give the answers to 3 significant figures:

A) NO2

B) glucose, C6H12O6

C) sodium hydroxide D) magnesium hydroxide

A. NO2 1 x AM of N = 14.00672 x AM of O = 2 x 15.9994 = 31.9988 MM of NO2 = 46.0055 46.0 amu

B. C6H12O6 6 x AM of C = 6 x 12.011 = 72.066 12 x AM of H = 12 x 1.0079 = 12.0948 6 x AM of O = 6 x 15.9994 = 95.9964 MM of C6H12O6 = 180.1572 180. amu

C. NaOH 1 x AM of Na = 22.98977 1 x AM of O = 15.9994 1 x AM of H = 1.0079 MM of NaOH = 39.9971 40.0 amu D. Mg(OH)2 1 x AM of Mg = 24.305 2 x AM of O = 2 x 15.9994 = 31.9988 2 x AM of H = 2 x 1.0079 = 2.0158 MM of Mg(OH)2 = 58.3196 58.3 amu

Avogadro

One mole of a substance is the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12g of C-12.

One mole of carbon = 6.02 x 1023 molecules of carbon

Avogadro’s Number (NA = 6.022 x 1023 molecules) is the numerical value assigned to the unit, 1 mole

Avogadro’s Number

Li2SO4

1. How many Li2SO4 units are in 3 moles of Li2SO4?

2. How many Li atoms are in 3 moles of Li2SO4?

3. How many S atoms are in 3 moles of Li2SO4?

4. How many O atoms are in 3 moles of Li2SO4?

Molar Mass of a substance is the mass of one mole of that substance

C-12 has a molar mass of exactly 12g/mol

For all substances, the molar mass (g/mol) is numerically equal to the formula mass (amu).

For example:1 mole O contains 6.02 x 1023 oxygen atoms1 mole of O2 contains 2x (6.02 x 1023) oxygen atoms

Molar Mass

Exercise 3.3What is the mass in grams of a calcium atom?

What is the mass in grams of an ethanol molecule?

23

1 mol

6.022 x 10 atoms

The atomic mass of Ca = 40.08 amu; thus, the molar mass = 40.08 g/mol, and 1 mol Ca = 6.022 x 1023 Ca atoms.Mass of one Ca =

= 6.6556 x 10−23

= 6.656 x 10−23 g/atom

40.08 g

1 mol Cax

2 6

46.07 g

1 mol C H O

The molecular mass of C2H5OH, or C2H6O, = (2 x 12.01) + (6 x 1.008) + 16.00 = 46.068. Its molar mass = 46.07 g/mol, and 1 mol = 6.022 x 1023 molecules of C2H6O.Mass of one C2H6O =

=

= 7.6503 x 10−23

= 7.650 x 10−23 g/molecule

x

Molar Mass

1 mol 6.022 x 1023 molecules

Mole Calculations

Consider ethanol with a molar mass of 46.1g/mol. Conversion to moles of ethanol:

1 mol C2H5OH = 46.1g/mol

Prepare acetic acid from 10.0g of ethanol. How many moles of ethanol is this?

10.0g C2H5OH x 1 mol C2H5OH = 0.217 mol C2H5OH 46.1 g C2H5OH

Mole Calculations

Exercise 3.4Dilute aqueous hydrogen peroxide is a colorlessliquid used as a bleach. Analysis of a solution shows it contains 0.909 mol of H2O2 in 1.00L of solution. What is the mass of hydrogen peroxide in this volume of solution?

The molar mass of H2O2 is 34.02 g/mol. Therefore,

2 22 2 2 2

2 2

34.02 g H O0.909 mol H O x = 30.92 = 30.9 g H O

1 mol H O

Mole Calculations

Exercise 3.5HNO3 is a corrosive liquid used for the manufacture of nitrogen fertilizers and explosives. A sample containing 28.5g of nitric acid was poured into a beaker. How many moles of HNO3 are there?

The molar mass of HNO3 is 63.01 g/mol. Therefore,

33 3

3

1 mol HNO28.5 g HNO x = 0.4523 = 0.452 mol HNO

63.01 g HNO

Mole Calculations

Exercise 3.6HCN is a volatile colorless liquid. The compound is highly poisonous . How many molecules are there in 56mg HCN, the average toxic does?

1 g

1000 mg

1 mol HCN

27.02 g HCN

236.022 x 10 HCN molecules

1 mol HCN

Convert the mass of HCN from milligrams to grams. Then convert grams of HCN to moles of HCN. Finally, convert moles of HCN to the number of HCN molecules.

56 mg HCN x

= 1.248 x 1021 = 1.2 x 1021 HCN molecules

x x

Chemical Formulas

• Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass.

Mass % A = mass A in the whole X 100%

whole mass

Exercise 3.7Ammonium nitrate, which is prepared from

nitric acid isused as a nitrogen fertilizer. Calculate the mass

percentagesof the elements in ammonium nitrate to 3 sig.

figs.

The molecular mass of NH4NO3 = 80.05; its molar mass = 80.05g/mol. Hence

Percent N = 28.02g N x 100% = 35.00 = 35.0%

80.05g NH4NO3 Percent H = 4 .032g H x

100% = 5.036 = 5.04% 80.05g NH4NO3

Percent O = 48.00g O x 80.05g NH4NO3

100% = 59.96 = 60.0%

Exercise 3.8How many grams of nitrogen are there in a fertilizer containing 48.5g of NH4NO3?

From the previous exercise, NH4NO3 is 35.0% N (Fraction N = 0.350), so the mass of N in 48.5 g of NH4NO3 is:

48.5 g NH4NO3 x 0.350 g N = 16.975 =

17.0 g N 1 g NH4NO3

Exercise 3.9A 3.87 mg of ascorbic acid (vitamin C) gives 5.80mg of CO2 and 1.58mg of H2O on combustion. What is the % composition of this compound?

First convert the mass of CO2 to moles of CO2. Next, convert this to moles of C (1 mol CO2 is equivalent to 1 mol C). Finally, convert to mass of C, changing milligrams to grams first:

5.80 x 10−3 g CO2 x 1 mol CO2 x 1 mol C x 12.01g C = 1.583 x 10−3g C 44.01g CO2 1 mol CO2 1 mol C

1.583mg C x 100% = 40.9% C 3.87mg

1.58 x 10-3g H2O x 1 mol H2O x 2 mol H x 1.0079g H = 1.767 x 10-4g H

18.02g H2O 1 mol H2O 1 mol H

1.767 x 10-4g H x 100% = 4.57% H3.87 x 10 -3g

Now that we have the % composition of C & H, we cancalculate % O:

1.583mg C x 100% = 40.9% C 3.87mg

1.767 x 10-4g H x 100% = 4.57% H3.87 x 10 -3g

The mass percentage of O can be determined by subtracting the sum of the above percentages from 100%:

Percent O = 100.000% − (40.90 + 4.5658) = 54.5342 = 54.5% O

Chemical Formulas

• Empirical Formula: Determined from data about percent composition, tells only the smallest whole number ratio of atoms in a compound.

• Tells the ratio of numbers of atoms in the compound

• hydrogen peroxide

• Molecular formula is H2O2

• Empirical formula is HO

Chemical Formulas

Exercise 3.10A sample of compound weighing 83.5g contains 33.4g of sulfur. The rest is oxygen. What is the empirical formula?Convert the masses to moles that are proportional to the

subscripts in the empirical formula:

83.5g compound – 33.4g S = 50.1g O

33.4g S x 1 mol S = 1.0414 mol S 50.1g O x 1 mol O =

3.1312 mol O 32.07g S 15.9994g ONext, obtain the smallest integers from the moles by dividing each

by the smallest number of moles:

For O: 3.1312 = 3.01 (3 oxygen) For S: 1.0414 = 1.00 (1 sulfur) 1.0414 1.0414

The empirical formula is SO3

Chemical Formulas

Exercise 3.11Benzoic acid is a white crystalline powder use a food preservative. The compound contains 68.8% C, 5.0% H and 26.2% O by mass. What is the empirical formula?

For a 100.0-g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 26.2 g are O. Using the molar masses, convert these

masses to moles:

68.8g C x 1 mol C = 5.729 mol C 12.01g C

5.0g H x 1 mol H = 4.961 mol H 1.0079g H

26.2g O x 1 mol O = 1.638 mol O 15.9994 g O

5.759 = 3.49 C 4.961 = 3.03 H 1.638 = 1.00 O

1.638 1.638 1.638

Rounding off, we obtain C3.5H3.0O1.0. Multiplying the numbers by 2 gives whole numbers, for an empirical formula of C7H6O2.

• Molecular Formula: a multiple of the empirical formula which is obtained by summing the atomic masses of the atoms in the empirical formula.

Molecular mass = n x empirical formula mass

n = the number of empirical formula units in the molecule. The molecular formula is obtained by multiplying the subscripts of the empirical formula by n.

Exercise 3.12The % composition of acetaldehyde is 54.5% C, 9.2% H and 36.3% O. The molecular mass is 44 amu. What is the molecular formula of acetaldehyde?

For a 100.0-g sample of acetaldehyde, 54.5 g are C, 9.2 g are H, and 36.3 g are O. Using the molar masses, convert these masses to moles:

54.5g C x 1 mol C = 4.537 mol C 4.537 = 2.0 C 12.01g C 2.270

9.2g H x 1 mol H = 9.127 mol H 9.127 = 4.0 H 1.0079g H 2.270

36.3g O x 1 mol O = 2.270 mol O 2.270 = 1.0 O 15.994 g O 2.270

The empirical formula of acetaldehyde is C2H4O

The empirical formula of acetaldehyde is C2H4O. The empirical formula mass is 2C x 12 amu = 24 amu 4H x 1 amu = 4 amu 1O x 16 amu = 16 amu

44 amu

To find n = molecular mass = 44 amu = 1 empirical formula mass 44 amu

Therefore the molecular formula is the same as the empirical formula

Stoichiometry

• Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities involved in a chemical reaction.

Grams of A

Convert grams of A to moles

of A

Convert moles of A to moles

of BConvert

moles of B to grams

of B

Grams of B

Stoichiometry

Exercise 3.14Sodium is a soft reactive metal that instantly reacts withwater to give hydrogen gas and a solution of sodium

hydroxide. How many grams of sodium metal is needed to give 7.81g of hydrogen by this reaction:

Equation: 2Na + 2H2O H2 + 2NaOH

From this equation, 2 moles of Na corresponds to one 1 mole of H2. Therefore,

7.81 g H2 x 1 mol H2 x 2 mols Na x 22.9898g Na = 178.1 = 178 g Na

2.0158g H2 1 mol H2 1 mol Na

Stoichiometry

Exercise 3.15Zinc sulfide reacts with oxygen to give zinc oxide and

sulfurDioxide. How many kg of oxygen gas combines with 5.00 x103 g of zinc sulfide in this reaction?

2 ZnS + 3 O2 2 ZnO + 2 SO2

5.00 x 103g ZnS x 1 mol ZnS x 3 mols O2 x 31.999g O2

97.474g ZnS 2 mols ZnS 1 mol O2

= 2462g O2 x 1kg 1000g

= 2.46kg O2

Stoichiometry

• Limiting Reagent: The reactant or the reagent that is completely consumed in a chemical reaction

– I need to make a fruit salad that is 1/2 apples and 1/2 oranges. I have 10 apples but only 7 oranges. What is the limiting fruit? How many apples and oranges can I use?

Exercise 3.17Aluminum chloride is used as a catalyst in various industrial reactions. It is prepared from hydrogen chloride gas and aluminum metal shavings. A reaction vessel contains 0.15 mol Al and 0.35 mol HCl. How many moles of AlCl3 can be prepared from this mixture?

2 Al (s) + 6 HCl (g) 2 AlCl3(s) + 3 H2(g)

First, determine the limiting reactant by calculating the moles of AlCl3 that would be obtained if Al and HCl were totally consumed:

0.15 mol Al x 2 mol AlCl3 = 0.15 mol AlCl3 2 mol Al

0.35 mol HCl x 2 mol AlCl3 = 0.12 mol AlCl3 (HCl is the limiting reactant) 6 mol HCl

Because the HCl produces the smaller amount of AlCl3, the reaction will stop when HCl is totally consumed but before all the Al is consumed. The limiting reactant is therefore HCl. The amount of AlCl3 produced must be 0.1166, or 0.12 mol.

Stoichiometry

• Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting?

Li2O(s) + H2O(l) 2 LiOH(s)

Li2OMM = 29.88 g/mol

H2OMM = 18.02 g/mol

Stoichiometry

• Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield.

• Theoretical Yield: maximum amount of product that can be obtained by a reaction from the given amounts of reactants.

• Percent Yield: the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated).

% Yield = Actual yield X 100% Theoretical yield

Stoichiometry

Exercise 3.19New industrial plants for acetic acid react

liquidmethanol with carbon monoxide in the

presenceof a catalyst.

CH3OH(l) + CO (g) HC2H3O2(l)

15.0g of methanol and 10.0g of carbonmonoxide were placed in a reaction vessel.What is the theoretical yield of acetic acid. If

theactual yield is 19.1g, what is the percent

yield?

Stoichiometry

Exercise 3.19Convert grams of each reactant to moles of

acetic acid:15.0g CH3OH x 1 mol CH3OH x 1 mol HC2H3O2 = 0.468mol

HC2H3O2

32.04g CH3OH 1 mol CH3OH

10.0g CO x 1 mol CO x 1 mol HC2H3O2 = 0.357mol HC2H3O2

(limiting)

28.01g CO 1 mol CO

Thus, CO is the limiting reactant, and 0.0357 mol HC2H3O2 is obtained. The mass of product is

0.357mol HC2H3O2 x 60.05g HC2H3O2 = 21.44g HC2H3O2

(theoretical) 1 mol HC2H3O2

The percentage yield is 19.1g x 100% = 89.08 = 89.1%

21.44g

Stoichiometry

• CH2Cl2 is prepared by reaction of CH4 with Cl2 giving HCl. How many grams of CH2Cl2 result from the reaction of 1.85 kg of CH4 if the yield is 43.1%?

CH4(g) + 2 Cl2(g) CH2Cl2(l) + 2 HCl(g)

1.85kg CH4 x 1000 g x 1 mol CH4 x 1 mol CH2Cl2 = 115.62 mol CH2Cl2 1 kg 16 g 1 mol CH4

115.62 mol CH2Cl2 x 84 g = 9712.5 g or 9.71kg CH2Cl2 produced

1 mol

43.1% =

Stoichiometry

K2PtCl4 + 2 NH3 Pt(NH3)2Cl2 + 2 KCl

• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react:

a) which is limiting reagent? b) how many grams of the excess reagent

are consumed? c) how many grams of cisplatin are

formed? K2PtCl4 = 415.08 g/mol NH3 = 17.04 g/mol

Pt(NH3)2Cl2 = 300.06 g/mol

Percent Composition

What is glucoses, C6H12O6, empirical formula, and what is the percentage composition of glucose?

Percent Composition

• Saccharin has the molecular formula C7H5NO3S, what is its empirical formula and the percent composition?

Empirical Formula

• A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. The molar mass is 58.11 g/mol.

– What is the empirical formula and molecular formula?