Below is a video showing how to find the percent composition of Na3PO4. The steps are written out below the video. Work along with the video instead of just watching!
Example #1: Simple Percent Composition
Find the percent composition of each element in Na3PO4.
Answer:
Step 1- You must first find the molar mass of the compound so we add together 3 Na + 1 P + 4 O = 163.91 g/mol
Step 2 - You have to pick an element to start with, for this example I am choosing Na, so the formula will be
3Na / (Na3PO4) -> (3*22.99)/(163.91) *100 = 42.08% Na
In the formula, you take the mass of the element of interest (Na) and multiply it by how many you have (3) and then you divide by the mass of the whole compound (163.91). Since we are dealing with percentages, we have to multiply the answer by 100.
We are not done because the question asked for the percentage of each element. If there were only two elements you could take the 42.08% and subtract from 100% but we have three elements so we must do another calculation, this time with P...
1P/Na3PO4 -> (30.97)/(163.91) *100 = 18.89% P
We followed the same procedure as we did to find Na except we used the mass of P. Now that we have two of the three elements you can take 100 - 18.89 - 42.08 = 39.03% O This works because percentages always have to add up to 100. We could have just as easily done the same percent composition calculation except with the mass of oxygen and gotten the same answer...below I prove this to you!
4 O /Na3PO4 -> (4*16)/(163.91) = 39.05% (The numbers are off by .02% due to rounding of masses but this is acceptable!)
Example #2: Multi-Step Percent Composition
You have 52.31 grams of water. How many grams are due to hydrogen?
Answer: First you must find the formula of water, which is H2O.
This problem is different from Example #1 because we are given a starting mass but your first step will still be to find the percent composition of water. Since the problem only asks for hydrogen, you do not need to find oxygen's percent.
2H/H2O -> (2*1.00794)/(2*1.00794+16) *100 = 11.19% H This is the percent of hydrogen in water but not the answer to the question. The question asks how many grams are due to hydrogen to solve this we take our starting mass and multiply it by the percentage of water...
52.31 g * .1119 = 5.854 g H (<- 4 sig. figs.)
When you do this problem you have to change the % into a decimal (move the decimal two places).
Find the percent composition of each element in Na3PO4.
Answer:
Step 1- You must first find the molar mass of the compound so we add together 3 Na + 1 P + 4 O = 163.91 g/mol
Step 2 - You have to pick an element to start with, for this example I am choosing Na, so the formula will be
3Na / (Na3PO4) -> (3*22.99)/(163.91) *100 = 42.08% Na
In the formula, you take the mass of the element of interest (Na) and multiply it by how many you have (3) and then you divide by the mass of the whole compound (163.91). Since we are dealing with percentages, we have to multiply the answer by 100.
We are not done because the question asked for the percentage of each element. If there were only two elements you could take the 42.08% and subtract from 100% but we have three elements so we must do another calculation, this time with P...
1P/Na3PO4 -> (30.97)/(163.91) *100 = 18.89% P
We followed the same procedure as we did to find Na except we used the mass of P. Now that we have two of the three elements you can take 100 - 18.89 - 42.08 = 39.03% O This works because percentages always have to add up to 100. We could have just as easily done the same percent composition calculation except with the mass of oxygen and gotten the same answer...below I prove this to you!
4 O /Na3PO4 -> (4*16)/(163.91) = 39.05% (The numbers are off by .02% due to rounding of masses but this is acceptable!)
Example #2: Multi-Step Percent Composition
You have 52.31 grams of water. How many grams are due to hydrogen?
Answer: First you must find the formula of water, which is H2O.
This problem is different from Example #1 because we are given a starting mass but your first step will still be to find the percent composition of water. Since the problem only asks for hydrogen, you do not need to find oxygen's percent.
2H/H2O -> (2*1.00794)/(2*1.00794+16) *100 = 11.19% H This is the percent of hydrogen in water but not the answer to the question. The question asks how many grams are due to hydrogen to solve this we take our starting mass and multiply it by the percentage of water...
52.31 g * .1119 = 5.854 g H (<- 4 sig. figs.)
When you do this problem you have to change the % into a decimal (move the decimal two places).