Molecular Formula f.19

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Andrea- 3J
Posts: 56
Joined: Tue Nov 14, 2017 3:00 am

Molecular Formula f.19

Postby Andrea- 3J » Fri Apr 06, 2018 3:40 pm

F.19
Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19g/mol and a mass percent composition of 49.48%C, 28.85%N, and 16.48%O. What is the molecular formula of caffeine?
Can someone help me by providing a steps to how to do this?

ElizabethP1L
Posts: 59
Joined: Wed Nov 15, 2017 3:01 am

Re: Molecular Formula f.19

Postby ElizabethP1L » Fri Apr 06, 2018 4:07 pm

The first step is to determine the number of MOLES of each element to find the molar ratio, which will help us derive the empirical formula. The molecular formula can then be found using this formula.

Since the problem did NOT specify how many grams of each atom we have, we can go ahead and assume that there are just 100g of each since it's a easy number to work with.

This gives us 49.48g C, 5.19 g H, 28.85 N, and 16.48g O all of which add up to 100 g!

To get the moles we need to divide these grams by their respective molar mass that can be calculated using atomic weights in our periodic table!

49.48g/12.01 g= 4.11 mol C

5.19g/1.008 g= 5.14 mol H

28.85g/14 g N= 2.06 mol N

16.48g/ 16 g= 1.03 mol O

Next, we divide each of these moles by the smallest mole number from our calculations (in this case, 1.03 mol O).

4.11 mol C/1.03 mol= 4 mol C (approximately 3.99, but for our purposes we may round up)

5.14 mol H/ 1.03 mol= 5 mol H (approximately 4.99, but for our purposes we may round up)

2.06 mol N/ 1.03 mol= 2 mol N

1.03 mol O / 1.03= 1 mol O
NOTE: These are all very similar to our above calculations because we are essentially dividing by 1
* since all of our above calculations are basically whole numbers we no longer need to modify them. If we had 3.66 mol of N for example, we would need to multiple by a whole number that would render it close to a whole number (in this case 3.66*3 would produce about 11 moles). All other mole rations would likewise need to be multiplied by 3.

Using our above molar ratios we know the empirical formula is C4H5N2O

To find our molecular formula, we need to find the molar mass of our empirical formula (4*12+5*1+2*14+16= 97 g/mol) The molar mass of the empirical formula will ALWAYS be less than the molecular formula unless the empirical formula is the molecular formula in which case they'd be equal.

We then divide the molar mass of caffeine 194.19 g/mol (which HAD to be given!) by the molar mass of our empirical formula, 97 g/mol:
194.19/97= 2

This means we need to multiple each of our molar ratios in the empirical formula (C4H5N2O) by 2!

The molecular forumla is C8H10N4O2!

Hope this helped! :)

KristinaNguyen_1A
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

Re: Molecular Formula f.19

Postby KristinaNguyen_1A » Sat Apr 07, 2018 10:27 am

1) Find the mass percentage (which is given)
2) You can assume that it is out of 100g (the mass percentage can be changed to grams)
3) Conver masses to molecular ratio (ex. mass ratio for carbon: 49.48g/12.01g*mol^-1)
4) Divide each by the smallest mol from the previous step and if needed, multiply with a number afterward to make them all whole numbers.
These steps will give you the empirical formula
To find the molecular formula:
5) You will need to be given a molar mass
6) Calculate the molar mass of the empirical formula
7) Find the ratio: (Given molar mass)/(empirical formula molar mass)
8) Multiply the value to the empirical formula to get the molecular formula


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