Chemistry...?

Question

Do you know how to find empirical and molecular formulas? I have a semester test tomorrow, and I have no clue what to do. All I know is that you need to know the empirical to find the molecular. Please help.

Ex.: Ribose has a molecular mass of 150 g/mol and a chemical comp of 40% C, 6.67% H, and 53.3% O. What is the molecular formula?

Ex.: A compound was analyzed in a lab to determine its empirical formula. It was determined to be 9 g C, 16.8L H, and 2.8L O. What is the molecular formula if the molar mass is 116g/mol?

Answer 1

First ex:
First, find how much mass of each atom is present on ribose:
C: 150g x 40% / 100 = 60g
O: 150g x 53.3% / 100 = 80g
H: 150g x 6.67% / 100 = 10 g

Then, convert the mass to moles, using the molar mass of each atom:

C: 60 g / 12 g/mol = 5 mol
O: 80 g / 16 g/mol = 5 mol
H: 10 g / 1 g/mol = 10 mol

If you want the empirical formula, divide each number of moles by the number of moles of the less abundant element (in this exercise, C or O):

C: 5mol / 5mol = 1
O: 5mol / 5mol = 1
H 10mol / 5 mol = 2

Then, the empirical formula for ribose is CH2O
To get the molecular formula, multiply the empirical formula by a factor that gives the right molecular mass of the molecule:

CH2O has a molecular mass of 30

150 / 30 = 5

So, the molecular formula of ribose is CH2O x 5 = C5H10O5


Second ex:
First, use the densities of O and H to find their masses:
O: 2,8L x 1,4 g/L = 3,92 g
H: 16,8L x 0,1 g/L = 1,68 g

Now proceed like the previous exercise:

C: 9 g / 12 g/mol = 0.75 mol
O: 3,92 g / 16 g/mol = 0.25 mol
H: 1.68 g / 1 g/mol = 1.68 mol

The less abundant element is O, so:

C: 0.75 / 0.25 = 3
O: 0.25 / 0.25 = 1
H: 1.68 / 0.25 = 6,72 so round to 6

The empirical formula is C3H6O
This formula gives a molecular mass of 58 g/mol, then:

116/58 = 2

Finally, to get the molecular formula of the unknown substance:
C3H6O x 2 = C6H12O2

Answer 2

You know that 1 mole has a mass of 150 g. This is the molecular weight of ribose.

You know that a quamtity of ribose (let's use 150 g) contains-
40% C or .40 x 150 = 60 g
6.67% H or 0.0667 x 150 = 10 g
53.3% O or 0.533 x 150 = 80 g

Since grams / atomic weight = moles of elements,
60 g of carbon = 60 / 12 = 5 moles of C
10 g of hydrogen = 10 / 1 = 10 moles of H
80 g of oxygen = 80 / 16 = 5 moles of O

The molecular formula was C5H10O5

What if you were told a sample was analyzed and found to contain 12 grams of carbon, 2 grams of hydrogen and 16 grams of oxygen. You would concluded that the empirical formula was CH2O but you wouln't know what the molecular formula was unless you knew the molar mass of the compound. CH2O = 12 + 2 + 16 = 30. If you were told the molar mass was 150, or 5 times that of CH2O, you would know the molecular formula was C5H10O5 (5 times CH2O).

Answer 3

OK Heather, don't panic!

For the first one, initially, you need to determine the ratio of the different elements in the molecule (this is not necessarily the formula, just the relative quantities of elements)

Here's how to do it:

Take the weight% of each element and divide by the atomic weight of that element; this will now give the ratio of elements by numbers of atoms (instead of weight)

40%C/12 = 3.33
6.67%H/1 = 6.67
53.3%O/16 = 3.33

We are not quite there yet; we need to take everything and divide it by the smallest of the set of numbers; this will give the ratio in the smallest possible whole numbers.

This gives the ratio of 1:2:1 for C,H and O respectively

We now have the emperical formula: CH2O!

Now we need to find the molecular formula. The weight of the empirical formula is: 12 + 2 + 16 or 30; we just need to see how many empirical formulas we need to add up to the given molecular weight of 150: Note: this should come out to exactly, or very nearly, a whole number, or there is a mistake somewhere!

150/30 = 5 so we need 5 "empirical formulas worth of atoms"

or C5H10O5 this is the molecular formula.

Now for the second problem, this is similar, but not exactly, so be careful.

The material was found to contain 9g of carbon; 16.8 liters of hydrogen and 2.8 liters of oxygen.

To get emperical formula we need to do this:

Carbon is 12 g/mole so we have 9/12 (or 0.75 mole of carbon.

We have 16.8l of hydrogen (remember that a mole of any gas at STP is 22.4liters)

[In this question they didn't specifically state that the gas volumes were measured at STP, but I'm giving them the benefit of the doubt]. (If they ever give you an equation like this where they state specifically that the gas volumes were taken at conditions other than STP, you can use the ideal gas law to get volumes at STP).

So 16.8/22.4 = 0.75 mole hydrogen

likewise 2.8/22.4 = 0.125 mole of oxygen

As before we get a ratio which we convert to the smallest whole numbers:

0.75 mole carbon
0.75 mole hydrogen (gas)
0.125 mole oxygen (gas)

0.75/0.125 = 6, so it would, at first, appear to be C6H6O but, THIS PART IS VERY IMPORTANT! 0.75 mole of hydrogen gas (H2) really contains 1.5 moles of H atoms! Likewise for the oxygen; There is realy twice as much hydrogen (and oxygen) as you would first expect! so the formula actually is C6H12O2. This formula just happens to weigh 116 amu.

I would suggest re-reading the section in your text book and how they do this, they may have a different procedure that is easier for you to understand.

I hope I haven't caused further confusion!

Good luck on your test! You won't have anyone on line to help you!