To compute for the empirical formula, assume there is 100 grams of the compound. That means there is 40.28 g B, 52.2 g N and 7.53 g H. Convert the mass into moles using their molar masses:
40.28 g B * 1 mol/10.811 g = 3.725835 mol B
52.2 g N * 1 mol/14 g = 3.72857 mol N
7.53 g H * 1 mol/1 g = 7.53 mol H
Divide all the moles by the smallest amount which is 3.725835 mol.
B: 3.725835/3.725835 = 1
N: 3.72857/3.725835 = 1
H: 7.53/3.725835 = 2
Therefore, the empirical formula is BNH₂.
Answer:
Flourine
Explanation:
Oxidation number will be +1 for oxygen and -1 for Flourine
Limited Genetic variability
Answer:
Kindly check the attached picture for the diagram of the chemical compound.
Explanation:
So, the following parameters were given from the question above;
=> A triplet at 0.9 ppm and a quartet at 1.4 ppm, a singlet at 1.35 ppm. Now, the unknown compound has a molecular formula of C7H16O.
For a triplet at 0.9ppm, there are nine (9) atoms of hydrogen, for the quartet at 1.4ppm there are six(6) atoms of hydrogen and for the singlet at 1.35 ppm, the number of hydrogen atoms is one(1). Hence, the total number of hydrogen atoms = 16.
Therefore, number of bondings = [(2 × number of carbon atoms) + 2 - number of hydrogen atoms present on the compound)/2 .
Thus number of bonds =[( 2× 7) + 2 - 16] ÷ 2 = 0.
Hence, there is no double bond or ring in the compound.