Asked by Katie
How do you calculate the molar masses for Strontium Phosphate and Chromium(III)Nitride using significant figures correctly?
Answers
Answered by
DrBob222
I am looking these atomic masses up at this site:
http://environmentalchemistry.com/yogi/reference/molar.html
For Sr3(PO4)2
Sr = 87.62 x 3 = 262.86
P = 30.97376 x 2 = 61.94752
O = 15.9994 x 8 = 127.9952
Total = 452.80272
The first multiplication allows 4 s.f. and I like to add one to keep from making rounding error; therefore, we will call it 262.86 (with the last place questionable).
The P has 7 s.f. so all will go.
The O has 6 s.f. so that will be 127.9952 with the 2 questionable.
Now, for adding the least precise number is .86 (two places to the right of the decimal); therefore, we can't have more than two places to the right of the decimal in the sum. That makes it 452.80 and since the 6 in .86 was carried as an "extra" I would truncate that last zero to make it 452.8
http://environmentalchemistry.com/yogi/reference/molar.html
For Sr3(PO4)2
Sr = 87.62 x 3 = 262.86
P = 30.97376 x 2 = 61.94752
O = 15.9994 x 8 = 127.9952
Total = 452.80272
The first multiplication allows 4 s.f. and I like to add one to keep from making rounding error; therefore, we will call it 262.86 (with the last place questionable).
The P has 7 s.f. so all will go.
The O has 6 s.f. so that will be 127.9952 with the 2 questionable.
Now, for adding the least precise number is .86 (two places to the right of the decimal); therefore, we can't have more than two places to the right of the decimal in the sum. That makes it 452.80 and since the 6 in .86 was carried as an "extra" I would truncate that last zero to make it 452.8
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