Empirical and

Molecular Formulas

How to find out what an unknown compound is

A chemist obtains a new product

What is the formula for the compound?

First step – determine constituent elements and their amounts

This info can be used to determine chemical formula

Formula of a compound represents relative numbers of atoms present

E.g. “CO2” tells us that in a molecule of this compound there is 1 carbon atom to every 2 oxygen atoms

To determine the formula of a substance we need to count the atoms – we can do this by weighing

An example

You have a compound that you know contains only Carbon, Hydrogen and Oxygen.

You have a 0.2015 g sample Analysis shows you have 0.0806 g C,

0.01353 g H, and 0.1074 g O We can convert these masses to

moles, and then moles to atoms using dimensional analysis…..

Determining # of moles of elements in unknown compound

Determining number of atoms of unknown compound

To summarize up to this point

We have .00671 moles of Carbon We have .01342 moles of Hydrogen We have .006713 moles of Oxygen

Amount of Carbon = Amount of Oxygen .01342/.006713 = 2, so we have twice

as much Hydrogen as we have of Oxygen and as we have of Carbon

We have a ratio of 1:2:1

We know the ratio of elements

1 Carbon : 2 Hydrogens : 1 Oxygen

We can write this as CH2O

Is this the molecular formula? Maybe…but the molecule might also have 2 Carbons, 4 Hydrogens and 2 Oxygens, or it might have 16 Carbons, 32 Hydrogens and 16 Oxygens

We have found the EMPIRICAL FORMULA – a formula that represents the ratio of elements in a compound.

This is also called “simplest formula” since it is smallest whole-number ratio of elements in the compound

Empirical Formula Vs Molecular Formula

Empirical formula gives relative numbers of atoms e.g. CH2O

Molecular formula gives the actual numbers of atoms e.g. C6H12O6

C6H12O6 = (CH2O)6

Find the Empirical Formula

Empirical Formula of Benzene = CH

Empirical Formula of Dioxin = C6H2Cl2O

How to calculate empirical formula

We have 0.2636 grams of nickel. We heat it in the presence of oxygen to produce 0.3354 grams of a nickel oxide. What is the formula of the compound we made?

First - what is the mass of oxygen that reacted with the copper?

Mass of copper oxide – Mass of copper = mass of oxygen

0.3354 g copper oxide – 0.2636 g copper = 0.0718 g oxygen

Next – Find number atoms involved…

Find number of atoms in the compound

•Mole quantities represent number of atoms•We have same number of moles of oxygen as of nickel•Empirical formula will be NiO

Another example

You have a metal oxide made by reacting 4.151g Al with 3.692 g O. What is the empirical formula?

Atomic mass Al = 26.98 g/mol

Atomic mass O = 16.00 g/mol

Need to know relative numbers of atoms, so need to convert grams to moles and then find the whole number ratio of atoms

Find whole number ratiosDo this by dividing both numbers by the smallest of the two. This converts the smallest number to 1

This is not a whole number ratio – to get to a whole number, all we have to do is multiply by 2

Finally we have found our empirical formula

Al2O3

Summary

If you have percent composition your can also find the empirical formula

You have a compound that is 27% Carbon and 73% Oxygen by mass. What is the empirical formula?

Assume you have 100 grams. That means you would have 27 g C and 73 g O

Convert grams to moles: 27 g C (1 mol/12 g C) = 2.25 mol C 73 g O (1 mol/16 g O) = 4.6 mol O Divide by smallest # of moles:

4.6 /2.25 = 2.04 (O) 2.25/2.25 = 1 (C)

Empirical formula = CO2

A poem to help

Percent to massMass to moleDivide by smallMultiply ‘til whole

Find empirical formula from percent composition

Nylon-6 is a compound that is 63.68 C, 12.38% N, 9.80% H and 14.4% O. Find the empirical formula

C6H11NO

One more step to find molecular formula

To find the molecular formula of your unknown compound, you need another piece of information – the molar mass of the compound, in addition to the percent composition

You have the empirical formula – what is the molecular formula?

Finding molecular formula

The molar mass of the compound will be some multiple of the empirical mass – find the multiple by dividing molar mass by empirical mass:

(238.88 g/mol)/(141.94 g/mol) = 1.7 1.7 ≈ 2 Molar mass of unknown is about twice

empirical mass. We will multiply empirical formula subscripts by 2

Molecular formula = P4O10

One last problem

Homework

Read Ch 6, section 3 (pp. 196 – 208) Do empirical formula homework sheets Do molecular formula homework sheets Monday – we will go over homework

and talk about hydrates Test Review Test on Wednesday!!!! We have 12 class meetings to go until

semester final