Calculate the empirical formula the compound with the following percent composition: 27.59%C 1.15%H 16.09%N 55.17%O

C2HNO3

Explanation: C = 27.59/12.011 = 2.297

H = 1.15/1.008 = 1.1409

N = 16.09/14.007 = 1.1487

O = 55.17/15.999 = 3.4483

Divide by smallest result:

C = 2

H=1

N=1

O = 3

Empirical formula = C2HNO3

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )

c. is the answer

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