What is a Mole?
We already covered what moles are in the Intro to Moles section, but to refresh your memory. A mole, or mol, is a standard unit for measuring large quantities of very small things such as atoms, molecules, or other types of particles.
The value of a mole of any substance contains 6.02x10^-3 molecules. This is called Avogadro's number.
Say for instance you wanted to find how many molecules were in 15 moles. You would set up the equation with the fence method that we used for the Unit Conversion Section in the Measurements and Problem Solving Section.
15 moles| 6.02x10^-3 = 9.03x10^24
| 1 mole
You get 9.03x10^24, but don't forget sig figs! Make sure you round to sig figs and your answer should be 9.0x10^24 molecules.
What if you wanted to find how many moles were in 7.5x10^19 molecules? That's simple. It's the same process, except this time Avogadro's number is on the bottom of the fence, and the mole is on the top of the fence like this.
7.5x10^19| 1 mole = 1.24x10^-4 moles
| 6.02x10^23
You get 1.24x10^-4 moles, but with sig figs your answer is 1.2x10^-4 moles.
Always remember sig figs! They are SO IMPORTANT!!
We already covered what moles are in the Intro to Moles section, but to refresh your memory. A mole, or mol, is a standard unit for measuring large quantities of very small things such as atoms, molecules, or other types of particles.
The value of a mole of any substance contains 6.02x10^-3 molecules. This is called Avogadro's number.
Say for instance you wanted to find how many molecules were in 15 moles. You would set up the equation with the fence method that we used for the Unit Conversion Section in the Measurements and Problem Solving Section.
15 moles| 6.02x10^-3 = 9.03x10^24
| 1 mole
You get 9.03x10^24, but don't forget sig figs! Make sure you round to sig figs and your answer should be 9.0x10^24 molecules.
What if you wanted to find how many moles were in 7.5x10^19 molecules? That's simple. It's the same process, except this time Avogadro's number is on the bottom of the fence, and the mole is on the top of the fence like this.
7.5x10^19| 1 mole = 1.24x10^-4 moles
| 6.02x10^23
You get 1.24x10^-4 moles, but with sig figs your answer is 1.2x10^-4 moles.
Always remember sig figs! They are SO IMPORTANT!!
Molar Mass
Molar mass is used in order to find the mass of a specific element, and the total mass of a compound. I really enjoy molar mass because I feel it's one of the easiest things to do in chemistry.
Say you're trying to find the molar mass of H2SO4.
First you need a periodic table that shows the atomic mass of every element, and then you look at the subscripts in your compound.
H has a subscript of 2, so go ahead and write this:
H: 2(
Then find H's atomic mass, which is 1.008. Plug that in.
H: 2(1.008)
S has no subscript so we can assume it is 1.
S: 1(
Find S's atomic mass, 32.07. Plug it in.
S: 1(32.07)
O has a subscript of 4.
O: 4(
Find O's atomic mass, 16.00. Plug it in.
O: 4(16.00)
Molar mass is used in order to find the mass of a specific element, and the total mass of a compound. I really enjoy molar mass because I feel it's one of the easiest things to do in chemistry.
Say you're trying to find the molar mass of H2SO4.
First you need a periodic table that shows the atomic mass of every element, and then you look at the subscripts in your compound.
H has a subscript of 2, so go ahead and write this:
H: 2(
Then find H's atomic mass, which is 1.008. Plug that in.
H: 2(1.008)
S has no subscript so we can assume it is 1.
S: 1(
Find S's atomic mass, 32.07. Plug it in.
S: 1(32.07)
O has a subscript of 4.
O: 4(
Find O's atomic mass, 16.00. Plug it in.
O: 4(16.00)
Now that each element is set up, you have to find their individual molar masses.
H: 2(1.008)= 2.016
S: 1(32.07)= 32.07
O: 4(16.00)= 64.00
After you get their individual masses, you have to add them together to find the molar mass of the compound.
2.016
+ 32.07
64.00
98.086 g
The molar mass of H2SO4 is 98.086 g
H: 2(1.008)= 2.016
S: 1(32.07)= 32.07
O: 4(16.00)= 64.00
After you get their individual masses, you have to add them together to find the molar mass of the compound.
2.016
+ 32.07
64.00
98.086 g
The molar mass of H2SO4 is 98.086 g
Percent Composition
The percent composition, or percentage composition, of a compound is a relative measure of the mass of each different element present in the compound.
To find the percent composition, you start off doing the same thing you did for molar mass. For the sake of saving time, we're going to use H2SO4 for this example too.
Remember you have to use the subscripts and atomic masses just like you did when you were trying to find molar mass.
H: 2(1.008)= 2.016
S: 1(32.07)= 32.07
O: 4(16.00)= 64.00
And you also have to add them together like you did molar mass.
2.016
+ 32.07
64.00
98.086 g
But now, you have to divide each element by the total.
H: 2(1.008)= 2.016/ 98.086= .0205
S: 1(32.07)= 32.07/ 98.086= .3269
O: 4(16.00)= 64.00/ 98.086= .6524
Then turn each new answer into a percentage.
H= 2.1%
S= 32.7%
O= 65.2%
The percent composition, or percentage composition, of a compound is a relative measure of the mass of each different element present in the compound.
To find the percent composition, you start off doing the same thing you did for molar mass. For the sake of saving time, we're going to use H2SO4 for this example too.
Remember you have to use the subscripts and atomic masses just like you did when you were trying to find molar mass.
H: 2(1.008)= 2.016
S: 1(32.07)= 32.07
O: 4(16.00)= 64.00
And you also have to add them together like you did molar mass.
2.016
+ 32.07
64.00
98.086 g
But now, you have to divide each element by the total.
H: 2(1.008)= 2.016/ 98.086= .0205
S: 1(32.07)= 32.07/ 98.086= .3269
O: 4(16.00)= 64.00/ 98.086= .6524
Then turn each new answer into a percentage.
H= 2.1%
S= 32.7%
O= 65.2%
An easy way to check yourself and see if you are right is to add up all the percent's and see if they equal to 100.
2.1
+32.7
65.2
100.0
2.1
+32.7
65.2
100.0
Empirical Formulas
Empirical, in chemistry, means "by experiment." Because of this, empirical formulas are formulas concocted from experimental data, rather than theory.
In order to solve an empirical formula, you have to be given the percent compositions of the compound.
A compound is 82.6% carbon, an 17.4% hydrogen by mass. Calculate the empirical Formula.
Well first you have to multiply each by 100, so that way it goes from a percent, to grams. Then you have to convert the grams to moles, using the fences. Use the periodic table to find how many grams of Carbon and Hydrogen there are, like what we did with the subscripts and atomic mass!
CH
C: 82.6g| 1 mole C = 6.87 moles C
| 12.01g C
H: 17.4g| 1 mole H = 17.26 moles H
| 1.008g H
Then take 6.87 and divide it by itself so you get 1, and then divide 17.26 by 6.87 to get 2.5. Round the 2.5 up to 3, and your empirical formula is CH3.
Empirical, in chemistry, means "by experiment." Because of this, empirical formulas are formulas concocted from experimental data, rather than theory.
In order to solve an empirical formula, you have to be given the percent compositions of the compound.
A compound is 82.6% carbon, an 17.4% hydrogen by mass. Calculate the empirical Formula.
Well first you have to multiply each by 100, so that way it goes from a percent, to grams. Then you have to convert the grams to moles, using the fences. Use the periodic table to find how many grams of Carbon and Hydrogen there are, like what we did with the subscripts and atomic mass!
CH
C: 82.6g| 1 mole C = 6.87 moles C
| 12.01g C
H: 17.4g| 1 mole H = 17.26 moles H
| 1.008g H
Then take 6.87 and divide it by itself so you get 1, and then divide 17.26 by 6.87 to get 2.5. Round the 2.5 up to 3, and your empirical formula is CH3.
Molecular Formulas
You have to have the Empirical Formula in order to solve the Molecular formula, and you also have to have the molar mass.
For the problem above, (CH3), the molar mass is 58g/mol. Calculate the molecular formula.
C: 1(12.01)
H: 3(1.008)
15.034
The mass of CH3 is 15.034. Now divide the molar mass by the empirical mass.
58 = 3.86
15.034
Round 3.86 up to 4. The molar mass is 4 times the empirical mass. Now take your previous formula, CH3 and mulitply it by 4. Your Molecular formula is C4H12.
You have to have the Empirical Formula in order to solve the Molecular formula, and you also have to have the molar mass.
For the problem above, (CH3), the molar mass is 58g/mol. Calculate the molecular formula.
C: 1(12.01)
H: 3(1.008)
15.034
The mass of CH3 is 15.034. Now divide the molar mass by the empirical mass.
58 = 3.86
15.034
Round 3.86 up to 4. The molar mass is 4 times the empirical mass. Now take your previous formula, CH3 and mulitply it by 4. Your Molecular formula is C4H12.