APPLICATIONS OF THE MOLE

APPLICATIONS OF THE

MOLE

Molar Mass of a Compound

• The molar mass of a compound is the mass of a mole of the representative particles of the compound.

• Because each representative particle is composed of two or more atoms, the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle.

To determine the molar mass of an element, find the element’s symbol on the periodic table and round the mass so there is one digit beyond the decimal.

The molar mass of carbon (C) is 12.0 g/mol, of chlorine (Cl) is 35.5 g/mol and of iron (Fe) is 55.8 g/mol.

Molar Mass of an Element


• In the case of NH3, the molar mass equals the mass of one mole of nitrogen atoms plus the mass of three moles of hydrogen atoms.


Molar mass of NH3 = molar mass of N + 3(molar mass of H)

Molar mass of NH3 = 14.0 g + 3(1.0 g) = 17.0 g/mol


• You can use the molar mass of a compound to convert between mass and moles, just as you used the molar mass of elements to make these conversions.

How many moles of magnesium in 56.3 g of Mg?

56.3 g Mg56.3 g Mg

g Mgg Mgmole Mgmole Mg

24.324.31____

(2.32)(2.32)

Molar Mass Conversion


• How many moles are in 146 grams of NH3?

(Answer: 8.59 mol)


• How many moles are in 295 grams of Cr(OH)3?

(Answer: 2.86 mol)


• How many moles are in 22.5 grams of HCl?

(Answer: 0.616 mol)

How many grams of sodium chloride in 3.45 moles of NaCl?

3.45 moles NaCl3.45 moles NaCl

mole NaClmole NaClg NaClg NaCl

1158.558.5____

(202)(202)



• How many grams are in 0.120 moles of AlF3?

(Answer: 10.1 g)


• How many grams are in 13.0 moles of H2SO4?

(Answer: 1275.3 g)


• How many grams are in 1.6 moles of K2CrO4?

(Answer: 311 g)

Percent Composition

• Recall that every chemical compound has a definite composition - a composition that is always the same wherever that compound is found.

• The composition of a compound is usually stated as the percent by mass of each element in the compound.

Percent Composition

• The percent of an element (X) in a compound can be found in the following way.

% X = (molar mass of X) (# X ‘s)

molar mass of compoundx 100

Example

• Determine the percent composition of chlorine in calcium chloride (CaCl2).

• First, analyze the information available from the formula.

• A mole of calcium chloride consists of one mole of calcium ions and two moles of chloride ions.

Calculating Percent Composition

• Next, gather molar mass information from the atomic masses on the periodic table.


• To the mass of one mole of CaCl2, a mole of calcium ions contributes 40.1 g, and two moles of chloride ions contribute 2 x 35.5 g = 71.0 g for a total molar mass of 111.1 g/mol for CaCl2.


• Finally, use the data to set up a calculation to determine the percent by mass of an element in the compound.

• The percent by mass of chlorine in CaCl2 can be calculated as follows.


% Cl = X 100

% X = (molar mass of X) (# X ions)


( g) ( Cl ions)

111.1

35.5

CaCl2

2

g

% Cl in CaCl2 = 63.9%

Example

• Determine the percent composition of carbon in sodium acetate (NaC2H3O2).

% X = (molar mass of X) (# X ions)



% C = X 100( g) ( C’s)

82.0

12.0

NaC2H3O2

2

g

% C = 29.3%

Calculate the percent composition aluminum of aluminum oxide (Al2O3).

Problem

52.9% Al

Determine the percent composition of oxygen in magnesium nitrate, which has the formula Mg(NO3)2.

Problem

64.7% O

Determine the percent composition of sulfur in aluminum sulfate, which has the formula Al2(SO4)3.

Problem

28.1% S

Determine the percent composition of oxygen in zinc nitrite, which has the formula Zn(NO2)2.

Problem

40.7% O

Percent Water in a Hydrate

• Hydrates are compounds that incorporate water molecules into their fundamental solid structure. In a hydrate (which usually has a specific crystalline form), a defined number of water molecules are associated with each formula unit of the primary material.


• Gypsum is a hydrate with two water molecules present for every formula unit of CaSO4. The chemical formula for gypsum is CaSO4 • 2 H2O and the chemical name is calcium sulfate dihydrate. Note that the dot in the formula (or multiplication sign) indicates that the waters are there.


• Other examples of hydrates are:

lithium perchlorate trihydrate - LiClO4 • 3 H2O;

magnesium carbonate pentahydrate - MgCO3 • 5 H2O;

and copper (II) sulfate pentahydrate - CuSO4 • 5 H2O.


• The water in the hydrate (referred to as "water of hydration") can be removed by heating the hydrate. When all hydrating water is removed, the material is said to be anhydrous and is referred to as an anhydrate.


• Experimentally measuring the percent water in a hydrate involves first heating a known mass of the hydrate to remove the waters of hydration and then measuring the mass of the anhydrate remaining. The difference between the two masses is the mass of water lost.


• Dividing the mass of the water lost by the original mass of hydrate used is equal to the fraction of water in the compound. Multiplying this fraction by 100 gives the percent water.

Example

• Determine the percent water in CuSO4 • 5 H2O (s).


• Mass of CuSO4 =

63.6 + 32.1 + 16.0 (4) = 159.7

• Mass of 5 H2O =

1.0 (2) + 16.0 = 18.0 x 5 = 90.0

• Mass of CuSO4 . 5 H2O =

159.7 + 90.0 = 249.7


% H2O = X 100( g) ( H2O’s)

249.7

18.0 5

g

% H2O = 36.0%

Problem

• Determine the percent water in MgCO3 • 5 H2O (s).

51.6% H2O

Problem

• Determine the percent water in LiClO4 • 3 H2O (s).

33.7% H2O

Empirical Formula

• You can use percent composition data to help identify an unknown compound by determining its empirical formula.

• The empirical formula is the simplest whole-number ratio of atoms of elements in the compound. In many cases, the empirical formula is the actual formula for the compound.

Empirical Formula

• For example, the simplest ratio of atoms of sodium to atoms of chlorine in sodium chloride is 1 atom Na : 1 atom Cl.

• So, the empirical formula of sodium chloride is Na1Cl1, or NaCl, which is the true formula for the compound.

Empirical Formula

• The formula for glucose is C6H12O6.

• The coefficients in glucose are all divisible by 6.

• The empirical formula of glucose is CH2O.

Problem

• Determine the empirical formula for Tl2C4H4O6.

TlC2H2O3

Problem

• Determine the empirical formula for N2O4.

NO2

Empirical Formula from Percent Composition

• The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound’s empirical formula.

• Because percent means “parts per hundred parts,” assume that you have 100 g of the compound.


• Then calculate the number of moles of each element in the 100 g of compound.


• The number of moles of manganese may be calculated as follows.

Given: 38.43% Mn

38.48 g Mn

54.9 g Mn

1 mol Mn= 0.7000 mol Mn


• The number of moles of carbon may be calculated as follows.

Given: 16.80% C

16.80 g C

12.0 g C

1 mol C= 1.400 mol C



Given: 44.77% O

44.77 g O

16.0 g O

1 mol O= 2.798 mol O


• The results show the following relationship.

mol Mn : mol C : mol O

0.7000 : 1.400 : 2.798


• To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles.

0.7000 mol Mn

0.7000= 1 mol Mn

• Find the whole number mole ratio for the compound.• These numbers become the subscripts in the empirical formula.


1.400 mol C

0.7000= 2 mol C

2.798 mol O

0.7000= 4 mol O


• The empirical formula for the compound is MnC2O4.

• The mole ratio for the compound is: 1 Mn : 2 C : 4 O

Example

• Determine the empirical formula of the following compound.

31.9 g Mg, 27.1 g P

Example

• The number of moles of magnesium may be calculated as follows.

Given: 31.9 g Mg

31.9 g Mg

24.3 g Mg

1 mol Mg= 1.31 mol Mg

Example

• The number of moles phosphorous may be calculated as follows.

Given: 27.1 g P

27.1 g P

31.0 g P

1 mol P= 0.874 mol P


• The results show the following relationship.

mol Mg : mol P

1.31 : 0.874


• To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles.

0.874 mol P

0.874= 1 mol P


1.31 mol Mg

0.874= 1.5 mol Mg


• You cannot have a fraction, so multiply both numbers by 2.

3 Mg : 2 P

• The mole ratio for the compound is: 1.5 Mg : 1 P


• The empirical formula for the compound is Mg3P2.

The composition of an unknown acid is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. Calculate the empirical formula for the acid.

Problem

CH20

The composition of an unknown ionic compound is 60.7% nickel and 39.3% fluorine. Calculate the empirical formula for the ionic compound.

Problem

NiF2

The composition of a compound is 6.27 g calcium and 1.46 g nitrogen. Calculate the empirical formula for the compound.

Problem

Ca3N2

Find the empirical formula for a compound consisting of 63.0% Mn and 37.0% O.

Problem

MnO2

Molecular Formulas • For many compounds, the empirical

formula is not the true formula.

• A molecular formula tells the exact number of atoms of each element in a molecule or formula unit of a compound.

Molecular Formulas

• The molecular formula for a compound is either the same as the empirical formula or a whole-number multiple of the empirical formula.

Molecular Formulas

• In order to determine the molecular formula for an unknown compound, you must know the molar mass of the compound in addition to its empirical formula.

Molecular Formulas

• Then you can compare the molar mass of the compound with the molar mass represented by the empirical formula as shown in the following example problem.

Example

• The molecular mass of benzene is 78 g/mol and its empirical formula is CH. What is the molecular formula for benzene?

Determining a Molecular Formula

• Calculate the molar mass represented by the formula CH.

• Here, the molar mass is the sum of the masses of one mole of each element.

Molar mass CH = 12.0 g + 1.0 g

Molar mass CH = 13.0 g/mol

Example

• As stated in the problem, the molecular mass of benzene is known to be 78.0 g/mol.


• To determine the molecular formula for benzene, calculate the whole number multiple, n, to apply to its empirical formula.

n =78.0 g/mol

13.0 g/mol= 6


• This calculation shows that the molar mass of benzene is six times the molar mass of its empirical formula CH.


• Therefore, the molecular formula must have six times as many atoms of each element as the empirical formula.

• Thus, the molecular formula is (CH)6 = C6H6

Example

• The simplest formula for butane is C2H5 and its molecular mass is about 60.0 g/mol. What is the molecular formula of butane?


• To determine the molecular formula for butane, calculate the whole number multiple, n, to apply to its empirical formula.

n =60.0 g/mol

29.0 g/mol= 2


• Therefore, the molecular formula must have two times as many atoms of each element as the empirical formula.

• Thus, the molecular formula is (C2H5)2 = C4H10

What is its molecular formula of cyanuric chloride, if the empirical formula is CClN and the molecular mass is 184.5 g/mol?

Problem

C3Cl3N3

The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molecular mass of vitamin C is about 180. What is the molecular formula of vitamin C?

Problem

C6H8O6

Example

• Maleic acid is a compound that is widely used in the plastics and textiles industries.

• The composition of maleic acid is 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen.

• Its molar mass is 116.1 g/mol. Calculate the molecular formula for maleic acid.


• Start by determining the empirical formula for the compound.

Example


Given: 41.39% C

41.39 g C

12.0 g C

1 mol C= 3.449 mol C

Example

• The number of moles of hydrogen may be calculated as follows.

Given: 3.47% H

3.47 g H

1.0 g H

1 mol H= 3.47 mol H

Example

• The number of moles of oxygen may be calculated as follows.

Given: 55.14% O

55.14 g O

16.0 g O

1 mol O= 3.446 mol O


• The numbers of moles of C, H, and O are nearly equal, so it is not necessary to divide through by the smallest value.

• You can see by inspection that the smallest whole-number ratio is 1C : 1H : 1O, and the empirical formula is CHO.


• Next, calculate the molar mass represented by the formula CHO.

• Here, the molar mass is the sum of the masses of one mole of each element.

Molar mass CHO = 12.0 g + 1.0 g + 16.0 g

Molar mass CHO = 29.0 g/mol

Example

• As stated in the problem, the molar mass of maleic acid is known to be 116.1 g/mol.


• To determine the molecular formula for maleic acid, calculate the whole number multiple, n, to apply to its empirical formula.

n =116.1 g/mol

29.0 g/mol= 4


• This calculation shows that the molar mass of maleic acid is four times the molar mass of its empirical formula CHO.


• Therefore, the molecular formula must have four times as many atoms of each element as the empirical formula.

• Thus, the molecular formula is (CHO)4 = C4H4O4



The composition of silver oxalate is 71.02% silver, 7.91% carbon, and 21.07% oxygen. If the molar mass of silver oxalate is 303.8 g/mol, what is its molecular formula?

Problem

Ag2C2O4

The composition of a compound is 85.6% carbon and 14.4% hydrogen. If the molar mass of the compound is 42.1 g/mol, what is its molecular formula?

Problem

C3H6

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APPLICATIONS OF THE MOLE