Chemical Composition Chapter 6

Chemical Composition Chapter 6

Tro, 2nd ed.

DEFINITIONS OF VARIOUS MASSESFormula or molecular mass = of atomic masses in

the chemical formula Molecular mass = mass in amu for a molecule, from

nonmetal elements forming covalent bondsMolecule is a covalent compound’s smallest unit,

made of nonmetals in covalent bondsFormula mass = mass in amu for a formula unit or for

all compoundsFormula unit is the smallest whole number ratio of an

ionic compound with metallic and nonmetallic elements in it (usually)

Molecular and formula mass

NaNO3 (NH4)2SO4

1 Na * 22.99 = 22.99 2 N * 14.01 = 28.02

1 N * 14.01 = 14.01 8 H * 1.008 = 8.064

3 O * 16.00 = 48.00 1 S * 32.07 = 32.07

85.00 amu 4 O * 16.00 = 64.00

132.15

amu

Molecular and formula mass

YOU PRACTICE:

PCl5 HNO3 CHLORINE GAS

ATOMS, MOLES & AVOGADRO’S NUMBER

The mass of a single atom is too small to measure on a balance.mass of hydrogen atom = 1.673 x 10-24 g

Chemists require a unit for counting which can express large numbers of atoms using simple numbers.

Chemists have chosen a unit for counting atoms.That unit is the

MOLE

ATOMS, MOLES & AVOGADRO’S NUMBER

1 mole = 6.0221 x 1023 objects6.0221 x 1023 objects/mole is Avogadro’s Number

6.0221 x 1023 is a very

number

LARGE

ATOMS, MOLES & AVOGADRO’S NUMBER

1 mole of any element contains 6.0221 x 1023 particles of that substance.The atomic mass in grams of any element contains 1

mole of atoms.This is the same number of particles as there are in

exactly 12 grams of carbon-12.

1 mole of atoms = 6.0221 x 1023 atoms1 mole of molecules = 6.0221 x 1023 molecules1 mole of ions = 6.0221 x 1023 ions1 mole of formula units = 6.0221 x 1023 formula units

AVOGADRO’S NUMBER & MOLAR MASS

The molar mass of an element is its atomic mass in grams INSTEAD of amu.

It contains 6.0221 x 1023 atoms (Avogadro’s number) of the element.

The units are grams/mole.

Element Atomic mass Molar massNumber of

atoms

H 1.008 amu 1.008 g 6.0221x1023

Mg 24.31 amu 24.31 g 6.0221x1023

Na 22.99 amu 22.99 g 6.0221x1023

Relating atomic mass and molar mass & Avogadro’s Number:

This is per atom. This is per mole. This is per mole.

THE MOLE MAP

MOLES

ATOMS, MOLECULES, FORMULA UNITS, IONS

MASS (grams)

VOLUME, mL

/NAX NA

X M / M

/ D X D

ATOMS, MOLECULES& AVOGADRO’S NUMBER

Practice: 1. How many atoms in 0.100 moles of He? 2. How many moles are represented by 9.00 x 109 atoms of He?

1. 0.100 mol * 6.0221 x 1023 atoms/mol = 6.0221 x 1022 He atoms

2.9.00x109atoms = 1.50 x 10-14 mol of He6.0221 x 1023 atoms/mol

OR9.00x109atoms * 1 mole = 1.50 x 10-14 mol He

6.0221 x 1023 atoms

MOLAR MASS – JUST LIKE CALCULATING MOLECULAR MASS1. Find molar mass of liquid bromine, Br2

2. Find molar mass of ethanol CH3CH2OH (organic structural formula)

3. Find molar mass of ammonium phosphate.

1. 2 * 79.90 g/mol = 159.80 g/mol

2. 2 C * 12.01 = 24.02 g/mol6 H * 1.008 = 6.048 g/mol1 O * 16.00 = 16.00 g/mol

46.068 = 46.07 g/mol (sig figs)

3. 149.12 g/mol

MASS, MOLECULES & THE MOLE: USING THE MOLE MAPEXAMPLE: Use mole map to find number of

molecules in 0.4607 grams of ethanol.First use molar mass to go from mass to moles:0.4607 * 1 mol = 0.01000 MOLES

46.07 gThen use moles and Avogadro’s Number to find

number of molecules:0.1000 mol * 6.0221 x 1023 molecules/mol

= 6.022 x 1021 molecules

MASS, MOLECULES & THE MOLE: USING THE MOLE MAPEXAMPLE: Use mole map to find number of

grams of lead in 9.00 x 1018 atoms of lead.First use Avogadro’s Number to find moles:

9.OO X 1018 atoms * 1 mol = 1.494 X 10-5 mol

6.0221 * 1023 atoms

Then use molar mass to find grams:

1.494 X 10-5 mol Pb * 207.2 g/mol = 3.10 X 10-3 g

(back to 3 sig figs)

COMPOUNDS, MOLAR MASS& AVOGADRO’S NUMBER

Chemical formulas are written to tell us:

- the number of atoms of each element

- the type of compound (ionic, covalent, acid, or organic)

- the structure of the compound

COMPOUNDS, MOLAR MASS& AVOGADRO’S NUMBER

NaNO3 (NH4)2SO4 PCl5 HNO3

1 Na atom 2 N atoms 1 P atom 1 H atom1 N atom 8 H atoms 5 Cl atoms 1 N atom3 O atoms 1 S atom 3 O atoms

4 O atoms

ionic ionic covalent acidformulaformula molecule moleculeunit unit (until aq)

H Cl HCl

6.0221 x 1023 H atoms

6.0221 x 1023 Cl atoms

6.0221 x 1023 HCl molecules

1 mol H atoms 1 mol Cl atoms1 mol HCl molecules

1.008 g H 35.45 g Cl 36.46 g HCl

1 molar mass H atoms

1 molar mass Cl atoms

1 molar mass HCl molecules

These relationships are present when hydrogen combines with chlorine.

2(2.00 mol O )23

2

2

6.022 x 10 molecules O1 mol O

2

2 atoms O1 molecule O

Conversion sequence: moles O2 → molecules O2 → atoms O

232

2

6.022 x 10 molecules O1 mol O

How many oxygen atoms are present in 2.00 mol of oxygen molecules?

Two conversion factors are needed:

2

2 atoms O1 mol O

24= 2.41 x10 atoms O

2

2

1 mol N28.02 g N

2(56.04 g N )23

2

2

6.022 x 10 molecules N

1 mol N

How many nitrogen atoms are in 56.04 g of N2?

The molar mass of N2 is 28.02 g/mol.

Conversion sequence: g N2 → moles N2 → molecules N2

→ atoms N

24= 2.409 x 10 atoms N2

2 atoms N1 molecule N

ELEMENTAL COMPOSITION

Elemental composition is the percent composition of a compound, which is the mass percent of each element in the compound

Example:

H2O is 11.19% H by mass & 88.81% O by masspercent = part * 100%

wholeElemental composition of sodium chloride, NaCl:1 Na*22.99= 22.99 g/mol (22.99/58.44)*100 = 39.34%1 Cl*35.45 = 35.45 g/mol (35.45/58.44)*100 = 60.66%

58.44 g/mol SUM = 100.00%

Percent Composition From FormulaIf the formula of a compound is known, a two-

step process is needed to calculate the percent composition.

Step 1 Calculate the molar mass of the formula.

Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.

Percent Composition From FormulaYou practice with ethanol: CH3CH2OH

2 C*12.01 = 24.02 g/mol (24.02/46.07)*100 = 52.14%

6 H*1.008 = 6.048 g/mol (6.048/46.07)*100 = 13.13%

1 O*16.00 = 16.00 g/mol (16.00/46.07)*100 = 34.73%

= 46.07 g/mol SUM = 100.00%

Percent Composition From Experimental Data

Percent composition can be calculated from experimental data without knowing the composition of the compound.

Step 1 Calculate the mass of the compound formed.

Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

Step 1 Calculate the total mass of the compound1.52 g N

3.47 g O

4.99 g

A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.

= total mass of compound

Step 2 Divide the mass of each element by the total mass of the compound formed.

3.47g O(100) = 69.5%

4.99g

1.52 g N(100) = 30.5%

4.99 g

N

30.5%

O

69.5%

Continued:

Mole Relationships inChemical Formulas

Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound

Moles of Compound Moles of Constituents

1 mol NaCl 1 mol Na, 1 mol Cl

1 mol H2O 2 mol H, 1 mole O

1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O

1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O

Mole Relationships inChemical Formulas

Class work: Skillbuilder 6.6

How many moles of oxygen are in 1.4 moles of sulfuric acid? How many grams of oxygen is this?

1.4 mol H2SO4 * 4 mol O/mol H2SO4 = 5.6 mol of oxygen

5.6 mol O * 16.00 g/mol = 89.6 grams of O

Empirical Formula versus Molecular Formula

The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.

The molecular formula is the true formula of a compound.

The empirical formula gives the relative number of atoms of each element present in the compound.

The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

Notice that these are all covalent compounds or diatomic elements. That’s because ionic formulas are by definition the lowest whole number ratio, with only a few exceptions like mercury (I) chloride, Hg2Cl2.

CalculatingEmpirical Formulas

Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if the actual amount is not given, and express the mass of each element in grams.

Step 2 Convert the grams of each element into moles of each element using each element’s molar mass.

Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value.– If the numbers obtained are whole numbers, use them

as subscripts and write the empirical formula.– If the numbers obtained are not whole numbers, go on

to step 4.

CalculatingEmpirical Formulas

Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers

Use these whole numbers as the subscripts in the empirical formula.

The results of calculations may differ from a whole number.If they differ ±0.05, round off to the next nearest whole

number.Deviations greater than 0.05 unit from a whole number usually

mean that the calculated ratios have to be multiplied by a whole number.

Some Common Fractions and Their Decimal Equivalents

Common FractionDecimal

Equivalent14

13

23

12

34

0.333…

0.25

0.5

0.75

0.666…

Resulting WholeNumber

1

1

2

1

3

Multiply the decimal equivalent by the number in

the denominator of the fraction to get a whole number.

The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.

Step 1 Express each element in grams. Assume 100 grams of compound.

K = 56.58 g

C = 8.68 g

O = 34.73 g

Continued:

Step 2 Convert the grams of each element to moles.

K: 56.58 g K1 mol K atoms

39.10 g K

1.447 mol K atoms

C: 8.68 g C1 mol C atoms

12.01 g C

0.723 mol C atoms

O: 34.73 g O1 mol O atoms

16.00 g O

2.171 mol O atoms

C has the smallest number of moles

0.723 mol C atoms

Continued:

Step 3 Divide each number of moles by the smallest value.1.447 mol

K = = 2.000.723 mol

0.723 molC: = 1.00

0.723 mol

2.171 molO = = 3.00

0.723 mol

The simplest ratio of K:C:O is 2:1:3

Empirical formula K2CO3

Now you try:

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

N2O5

Calculating the Molecular Formula from the Empirical

FormulaThe molecular formula can be calculated from the empirical formula if the molar mass is known.

The molecular formula will be equal to the empirical formula or some multiple, n, of it.

To determine the molecular formula evaluate n.

n is the number of units of the empirical formula contained in the molecular formula.

molar massn = =

massof empirical formula formula units number of empirical

What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2 g?

The molecular formula is (CH2)9 = C9H18

Calculate the mass of each CH2 “mol”

1 C = 1(12.01 g/mol) = 12.01g/mol

2 H = 2(1.01 g/mol) = 2.02g/mol

14.03g/mol126.2 g

n 9 (empirical formula units)14.03 g

Group Work:

What is the empirical formula and the molecular formula of lactic acid, which is 40.00% C, 6.71% H and 53.29% O, with a molar mass of 90.1 g?

C3H6O3 (or CH3CHOHCOOH)