Empirical (or simplest) Formula

What does empirical mean?

Determined by experiment.

What is the simplest whole number ratio of C : H in benzene, C6H6?

1 : 1What is the simplest whole number ratio of

C : H in acetylene, C2H2?1 : 1The empirical, (or simplest), formula for both

benzene and acetylene is C1H1. Is this the actual, (or molecular), formula of

each?No. Now look at these . . .

name molecular formula

empirical (simplest) formula

lowest ratio of elements

hydrogen peroxide

H2O2 HO

glucose C6H12O6 CH2O

acetylene C2H2 CH

benzene C6H6 CH

aniline C6H7N C6H7N

water H2O H2O

Can the molecular (actual) formula of a compound be the same as its simplest (empirical) formula?

Sometimes.

But not always.

Write an equation relating the mf of a compound to its sf. (ie. mf = ????)

mf = n*(sf),

where n = 1, 2, 3 . . .

How to determine the simplest formula of a compound

What is the sf of a cmpd that is 85.6% C and 14.4% H, by mass. (These will always be by mass.)

In 100 g of the compound, how many grams of C will there be?

85.6 g C. How many grams H?

14.4 g H

Now convert these masses to # moles . . .

mol C = 85.6 g C/12.01 g C·mol-1 (extra sf)

mol C = 7.127 (No rounding ‘til end.)

mol H = 14.4 g H/1.01 g H·mol-1

= 14.26

Now write mol : mol ratio of C : H.

7.127 mol C : 14.26 mol H Divide both by smallest # mol to get lowest ratio.

1.00 mol C : 2.00 mol H

sf of cmpd is C1H2 . . . but . . .

Is this the molecular formula of the compound?

No CH2 cannot exist. C needs to makefour bonds.You can also solve sf problems using a table:

simplest formula = C1H2

(More data req’d to determine the actual formula—stay tuned.)

Element % g/100g mm (g/mol)

# mol ratio

C 85.6 85.6 12.01 7.127 1

H 14.4 14.4 1.01 14.26 2

Try this one:A compound is 46.3% Li and 53.7% O by mass.

Determine the sf of this cmpd. Make a table.

sf = Li2O Does this formula make sense?

Yes. Li2O is an ionic compound.

Element % g/100g mm # mol ratio

Li 46.3 46.3 6.941 6.671 2

O 53.7 53.7 16.00 3.356 1

Homework

p 270 7 – 12 (do ‘em all)

p 273 31 – 40 (do a few)

So you think you’re so smart . . .Try this one. Make a table if you want.

The percentage composition of a fuel is 81.7% C and 18.3% H by mass. Find the sf of the fuel.

Can we round off C1H2.66 ?

No—we need to express 2.66 as a fraction.

C1H8/3 What can you multiply both subscripts by to get rid of the denominator?

Multiply by 3 sf = C3H8 When do we do this?

Element % g/100g mm # mol ratio

C 81.7 81.7 12.01 6.80 1H 18.3 18.3 1.01 18.11 2.66

When you see decimal multiply all subscripts by

XX.50 or ½ 2

XX.80 or 4/5 5

XX.75 or ¾ 4

XX.67 or 2/3 3

XX.60 or 3/5 5

XX.40 or 2/5 5

XX.33 or 1/3 3

XX.25 or ¼ 4

XX.20 or 1/5 5

XX.17 or 1/6 6

Try this one:

An inorganic salt is composed of 17.6% Na, 39.7% Cr, 42.8% O. Find the sf. Table!!!

ratio is Na1Cr1O3.5, but we need to multiply by

2 to get sf = Na2Cr2O7.

Element g/100g mm # mol ratio

Na 17.6 22.99 0.7655 1

Cr 39.7 52.00 0.763 1

O 42.8 16.00 2.675 3.50

Homework

p 275 46, 42, 47 (in this order + more if you want)