Empirical formula question?
i can do everything once i know the amount of mass of each substance in the compound, but i do not know how to get the mass % elements to grams of each element
thanks
Answer:
You make an assumption. I tell my students to assume 100. grams of the compound. That converts your percentages directly to grams:
87.5 g N
12.5 g H
Now, convert them to moles:
87.5 g N (1 mol/14.01 g) = 6.25 mol N
12.5 g H (1 mol/1.01 g) = 12.4 g H
Now, divide by the smallest # of moles:
6.25 g N/6.25 g = 1.00
12.4 g H/6.25 g = 1.98 = 2.00
NH2 is your empirical formula. This is nothing magical about 100 g, just very easy. You can choose any gram amount of the compound because you know the % breakdown. 1, 10 or 100 will be the easiest, but I could choose 63.4 and still get the correct answer. I would have to multiply the % by 63.4 to get the grams of N and H.
Please - what's the other element?
To convert from Mass % to empirical formula, you need to divide by the atomic mass of each element.
87.5% (mass) nitrogen / 14.000674(g Nitrogen/mol nitrogen) = ~6.25% mol nitrogen
do the same for the other elements
12.5% say oxygen, because you didnt list one.
12.5% / 15.9994g/mol = ~.78 % mol oxygen
then you need to establish a simple ratio (if one exists) between them. usually just divide them by each other and youll get something simple.
6.25/0.78 = 8 or 8:1
so the empiriccal formula for 87.5% nitrogen 12.5% oxygen would be
N8O im not sure that exists, but thats how to do it.
1. Take the percentages and divide by the atomic mass of the elements, (87.5% / 14 and 12.5%/ whatever the other one is)
2. You will get ratios 6.25: the other one, divide both sides by the smaller number (6.25 or the other one)
3. Roughly get to whole number ratios it may be something like 1:1.9999...etc so round to 1:2.
4. Than that should give you the empirical values.
It will help a bit more if you give us the other element's identity.
Assume you have 100 g of the substance. Then you have 87.5 g N and 12.5 g of the other element. Divide each mass of each element by its molar mass. (i.e. 87.5g N/(14 g)) This gives you the moles of each element. Divide by the lower number of moles. (i.e. if your calculations are 4.45 mol N and 3.22 mol of the other element, divide 4.45 by 3.22 and 3.22 by 3.22) You should end up with numbers that are pretty close to whole numbers (i.e you might need to round 3.97 up to 4 or 2.99 up to 3) but these are the numbers (i.e. 4, 3) of moles of each element in the empirical formula. So the emperical formula would be for example Nsub4Ysub3
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