1. A student calculates the empirical formula of a compound to be C1.5H3.5. Express this as a correct empirical formula.
2. The formula for acetic acid, an ingredient of vinegar, is C2H4O2. What is the empirical formula for acetic acid?
3. A compound contains 69.9% iron and 30.1% oxygen. What is its empirical formula?
If you could help me that'd be awesome. ^^ Thanks a lot!! :)
2. The formula for acetic acid, an ingredient of vinegar, is C2H4O2. What is the empirical formula for acetic acid?
3. A compound contains 69.9% iron and 30.1% oxygen. What is its empirical formula?
If you could help me that'd be awesome. ^^ Thanks a lot!! :)
-
1. A correct empirical formula uses whole number ratios between the elements. To get this, multiply C1.5H3.5 by 2, which is C3H7.
2. To get the smallest whole number ratio between the elements, divide C2H4O2 by 2, which is CH2O.
3. Based on the percentages given 100 g of the compound would contain 69.9 g Fe and 30.1 g O. Convert those to moles by dividing each mass by the atomic weight from the periodic table.
69.9 g Fe x (1 mole Fe / 55.85 g Fe) = 1.25 moles Fe
30.1 g O x (1 mole O / 16.00 g O) = 1.88 moles O
Dividing both of those by the smaller (1.25) gives us
Fe: 1.25 / 1.25 = 1.00
O: 1.88 / 1.25 = 1.50
Doubling those two gives us Fe2O3.
2. To get the smallest whole number ratio between the elements, divide C2H4O2 by 2, which is CH2O.
3. Based on the percentages given 100 g of the compound would contain 69.9 g Fe and 30.1 g O. Convert those to moles by dividing each mass by the atomic weight from the periodic table.
69.9 g Fe x (1 mole Fe / 55.85 g Fe) = 1.25 moles Fe
30.1 g O x (1 mole O / 16.00 g O) = 1.88 moles O
Dividing both of those by the smaller (1.25) gives us
Fe: 1.25 / 1.25 = 1.00
O: 1.88 / 1.25 = 1.50
Doubling those two gives us Fe2O3.
-
1. C3H7- The definition of an empirical formula is the amount of atoms in a compound expressed as the most simplified whole number ratio possible. In this specific case, 3 is multiplied to 1.5 and 3.5 to get the smallest and most simplified whole number ratio.
2. CH2O- I divided each number by 2 to get the smallest and most simplified whole number ratio.
3. Treat percentages as number of grams, and convert those grams to moles.
For Iron, you have 69.9 grams, and you must convert this amount to moles by using molar mass (69.9 divided by 55.8)...You have 1.25 moles of Iron.
For Oxygen, you have 30.1 grams, and you must convert this to moles by using molar mass (30.1 divided by 16.0)....You have 1.88 moles of Oxygen.
Now, you must divide each final mole number by the smallest mole number out of the two. In this case, you would divide each mole number by 1.25.
-For Iron: 1.25 divided by 1.25...You have 1 atom of Iron.
-For Oxygen: 1.88 divided by 1.25...You have 1.5 atoms of Oxygen.
Since 1.5 is not a whole number, you need to multiply this number by 2, as well as the Iron atom number.
-1 x 2= 2 atoms of Iron.
-1.5 x 2= 3 atoms of Oxygen.
The empirical formula is Fe2O3.
2. CH2O- I divided each number by 2 to get the smallest and most simplified whole number ratio.
3. Treat percentages as number of grams, and convert those grams to moles.
For Iron, you have 69.9 grams, and you must convert this amount to moles by using molar mass (69.9 divided by 55.8)...You have 1.25 moles of Iron.
For Oxygen, you have 30.1 grams, and you must convert this to moles by using molar mass (30.1 divided by 16.0)....You have 1.88 moles of Oxygen.
Now, you must divide each final mole number by the smallest mole number out of the two. In this case, you would divide each mole number by 1.25.
-For Iron: 1.25 divided by 1.25...You have 1 atom of Iron.
-For Oxygen: 1.88 divided by 1.25...You have 1.5 atoms of Oxygen.
Since 1.5 is not a whole number, you need to multiply this number by 2, as well as the Iron atom number.
-1 x 2= 2 atoms of Iron.
-1.5 x 2= 3 atoms of Oxygen.
The empirical formula is Fe2O3.