To what category of elements does an element belong if it is a poor conductor of electricity
1 answer:
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We have the next % composition:
C. 49.48%
H, 5.19%.
N. 28.85%
0. 16.48%
We assume 100 g of sample
1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.
C. 49.48 g
H, 5.19 g
N. 28.85 g
0. 16.48 g
2) We calculate the number of moles of each element (we need the mass per mole of each element)
For C) 12.01 g/mol
49.48 g x (1 mol/12.01 g) = 4.120 moles
For H) 1.007 g/mol
5.19 g x (1 mol/1.007 g) = 5.154 moles
For O) 15.99 g/mol
16.48 g x (1 mol/15.99 g) = 1.030 moles
For N) 14.00 g/mol
28.85 g x (1 mol/14.00 g) = 2.060 moles
3) We choose the smallest number from 2) and divide the rest of them by it.
For C) 4.120 moles/1.030 moles= 4
For H) 5.154 moles/1.030 moles= 5
For O) 1.030 moles/1.030 moles= 1
For N) 2.060 moles/1.030 moles= 2
4) The numbers in 3) represents the subindex from the empirical formula of caffeine:
5) We calculate the molar mass of our empirical formula, 97.06 g/mol.
We already have the molar mass of the molecular formula, so we proceed like this:
n= the molar mass of the molecular formula/the molar mass of the empirical formula
n = 194.19 g/mol/97.06 g/mol = 2 approx.
We use "n" and we multiply our empirical formula by n = 2:
Therefore, our molecular formula: