The Mole Concept & Problem Solutions

 

1. If atomic mass of Mg atom is 24 g, find mass of 1 Mg atom.

Solution:

We can solve this problem in to ways;

1^st way:

6,02x10²³ amu is 1 g

24  amu           is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=4x10^-23 g

2^nd way;

1 mol Mg (6,02x1023 Mg atoms) is 24 g

6,02x10²³ Mg atoms   24 g

1              Mg atom     ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=4x10^-23 g

 

2. Find mass of 1 molecule C₂H₆. (C=12, H=1)

Solution:

C₂H₆=2.12+6.1=30

6,02x10²³ C₂H₆ molecule is 30 g

1              C₂H₆ molecule is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

1 C₂H₆ molecule is =?=5.10^-23 g

 

3. Find mole of 6,9 g Na. (Na=23)

Solution:

23 is the atomic mass of Na, in other words 1 mole Na is 23 g.

23 g Na is 1 mol

6,9 g Na is ? mol

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,3 mol

 

4. Find mass of 0,2 mol P₄ . (P=31)

Solution:

Molecule mass of P₄ =4.31=124 g

1 mol P₄ is 124 g

0,2 mol P₄ is ? mol

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?= 24,8 g

 

5. Find mole of 4,48 liters O₂ under normal conditions.

Solution:

Under normal conditions, 1 mol gas is 22,4 liters. We use following formula to find moles of gas under normal conditions;

n=V/22,4

n=4,48/22,4=0,2 mol

 

6. Find mass of Fe in the compound including 4,8x10²³ O atoms ;Fe₃O₄ .

Solution:

We first find mole of O in the compound;

nO=(4,48x10²³)/(6,02x10²³)=0,8mol

Now we find mole of compound that contains 0,8mol O;

n_Fe3O4=(moles of O)/(moles of O in compound)=0,8/4=0,2mol

Mole of Fe in compound is;

There are 3 mol Fe in 1 mol Compound

there are ? mol Fe in 0,2mol compound

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,6 mol Fe in compound

mass of 0,6 mol Fe is;

1 mol Fe is 56 g

0,6 mol Fe is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯

?=33,6g There are 33,6 g Fe in compound

The Mole Concept Exam2 and Problem Solutions

The Mole Concept Exam2 and  Problem Solutions

 

1. Find relation between number of molecules of given matters;

I. C₂H₂ that includes 2mol H atom

II. CH₄ that includes N atoms (N is Avogadro number)

III. C₃H₄ that includes 1,5 N C atoms

Solution:

I.

1mol C₂H₄ includes 4mol H atom

?mol C₂H₄ includes 2mol H atom

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,5mol C₂H₄

II.

1mol CH₄ includes 5N atom

?mol CH₄ includes N atom

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,2mol CH₄

III.

1mol C₃H₄ includes 3N C atom

?mol C₃H₄ includes 1,5N C atom

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,5mol C₃H₄

Thus relation between them: I=III>II

 

2. Find relation between number of atoms of given matters.

I. 6 PH₃ molecules

II. CO₂ that includes 24N atom

III. 8mol O₃

Solution:

I.

1 PH₃ molecule contains 4 atoms

6 PH₃ molecules contain ? atom

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=24 atoms

II.

CO₂ includes 24 N atoms.

III.

1mol O₃ contains 3N atoms

8mol O₃ contain ?N atoms

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=24N atoms

 

3. Which one of the following statements is false for 0,5mol C₂H₆? (C=12, H=1 and take Avogadro Number=6x10²³)

I. It is 15 g

II. It includes 3x10²³ C₂H₆ molecules

III. It includes 1mol C atom

IV. It includes 4 atoms.

V. It includes 3 g H.

Solution:

I. molar mass of C₂H₆=2.(12) + 6.(1)=30g/mol

1mol C₂H₆ is 30g/mol

0,5mol C₂H₆ is ?mol

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=15 g/mol I is true

II.

1mol C₂H₆ includes 6x10²³ C₂H₆ molecules

0,5mol C₂H₆ includes ?C₂H₆ molecules

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=3x10²³ C₂H₆ molecules II is true

III.

1mol C₂H₆ includes 2mol C atoms

0,5mol C₂H₆ include ?mol C atoms

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=1mol C atom, III is true

IV.

1mol C₂H₆ includes 8N atoms

0,5mol C₂H₆ includes ?N atoms

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=4N atoms, IV is false.

V.

1mol C₂H₆ includes 6g H

0,5mol C₂H₆ includes ?g H

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=3 g H  V is true

 

4. If, 8,4 g X element includes 9,03x10²² atoms and 0,1mol X₂Y₃ compound is 16g find atomic mass of Y element. (Avogadro number is 6,02x10²³)

Solution:

Mole of X element is ;

nx=(9,03x10²²)/(6,02x10²³)=0,15mol

Atomic mass of X;

A_X=8,4/0,15=56g/mol

Molar mass of X₂Y₃ compound;

M_X2Y3=16/0,1=160g/mol

We find atomic mass of Y by;

2.X + 3Y =160

2.(56) + 3(Y)=160

Y=16g/mol is atomic mass of Y.

 

5. Which ones of the following statements are true for 0,2mol C₃H₄ and 0,5mol C₂H₆ gas mixture? (C=12, H=1 and N=Avogadro Number)

I. Mass of mixture is 23 g.

II. It includes 1,6mol C atom.

III. It includes 0,7N molecule.

Solution:

I. Molar mass of C₃H₄ =3.(12) + 4.(1)=40g/mol

1mol C₃H₄ is 40g

0,2mol C₃H₄ ?g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=8g C₃H₄

Molar mass of C₂H₆=2.(12) + 6.(1)=30g/mol

1mol C₂H₆ is 30 g

0,5 mol C₂H₆ is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=15 g C₂H₆

Total mass of mixture is=8 + 15=23 g I is true

II.

1mol C₃H₄ contains 3mol C atoms

0,2mol C₃H₄ contains ?mol C atoms

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,6mol C atoms

1mol C₂H₆ contains 2mol C atoms

0,5mol C₂H₆ contains ?mol Catoms

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=1mol C atom

Total umber of C atoms in mixture is=0,6 + 1=1,6mol C atoms. II is true.

III.

0,2mol C₃H₄ + 0,5mol C₂H₆ = 0,7mol molecule

There are 0,7N molecule in 0,7 mol mixture, III is true.

Mole-Mass and Mass-Mass Problems

Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.

We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:

As an example, consider the balanced chemical equation

Fe2O3 + 3SO3 → Fe2(SO4)3

If we have 3.59 mol of Fe2O3, how many grams of SO3 can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3 in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3 needed. Then using the molar mass of SO3 as a conversion factor, we determine the mass that this number of moles of SO3 has.

The first step resembles the exercises we did in . As usual, we start with the quantity we were given:

The mol Fe2O3 units cancel, leaving mol SO3 unit. Now, we take this answer and convert it to grams of SO3, using the molar mass of SO3 as the conversion factor:

Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3 will react with 3.59 mol of Fe2O3. Many problems of this type can be answered in this manner.

The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:

We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.

EXAMPLE 8

How many grams of CO2 are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation?

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

Solution

Our strategy will be to convert from moles of HCl to moles of CO2 and then from moles of CO2 to grams of CO2. We will need the molar mass of CO2, which is 44.01 g/mol. Performing these two conversions in a single-line gives 46.0 g of CO2:

 

The molar ratio between CO2 and HCl comes from the balanced chemical equation.

SKILL-BUILDING EXERCISE

How many grams of glucose (C6H12O6) are produced if 17.3 mol of H2O are reacted according to this balanced chemical equation?

6CO2 + 6H2O → C6H12O6 + 6O2

It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:

This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.

EXAMPLE 9

Methane can react with elemental chlorine to make carbon tetrachloride (CCl4). The balanced chemical equation is as follows:

CH4 + 4Cl2 → CCl4 + 4HCl

How many grams of HCl are produced by the reaction of 100.0 g of CH4?

Solution

First, let us work the problem in stepwise fashion. We begin by converting the mass of CH4 to moles of CH4, using the molar mass of CH4 (16.05 g/mol) as the conversion factor:

Note that we inverted the molar mass so that the gram units cancel, giving us an answer in moles. Next, we use the balanced chemical equation to determine the ratio of moles CH4 and moles HCl and convert our first result into moles of HCl:

Finally, we use the molar mass of HCl (36.46 g/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl:

In each step, we have limited the answer to the proper number of significant figures. If desired, we can do all three conversions on a single line:

This final answer is slightly different from our first answer because only the final answer is restricted to the proper number of significant figures. In the first answer, we limited each intermediate quantity to the proper number of significant figures. As you can see, both answers are essentially the same.

SKILL-BUILDING EXERCISE

The oxidation of propanal (CH3CH2CHO) to propionic acid (CH3CH2COOH) has the following chemical equation:

CH3CH2CHO + 2K2Cr2O7 → CH3CH2COOH + other products

How many grams of propionic acid are produced by the reaction of 135.8 g of K2Cr2O7?

CONCEPT REVIEW EXERCISES

What is the general sequence of conversions for a mole-mass calculation?

What is the general sequence of conversions for a mass-mass calculation?