Question

Q1. The hormone norepinephrine is released in the human body during stress and increases the body’s metabolic rate. Like many biochemical compounds, norepinephrine is composed of carbon, hydrogen, oxygen, and nitrogen. The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O, and 8.28% N. What is the empirical formula of norepinephrine?  

Q2. Elemental magnesium consists of three isotopes: ²⁴Mg with an accurate mass of 23.99 amu, ²⁵Mg with an accurate mass of 24.99 amu, and ²⁶Mg with an accurate mass of 25.99 amu. If ²⁴Mg is 79% of naturally occurring magnesium, what is the percent abundance of ²⁶Mg?   

A) 10%

B) 11%

C) 15%

D) 19%

E) 21%

Q3. A compound containing xenon and fluorine was prepared by shining sunlight on a mixture of Xe (0.526 g) and excess F₂ gas. If you isolate 0.678 g of the new compound, what is its empirical formula?  

Answer 1

Q1

Empirical formula --> the least mathematical ratio between species possible.

Assume a basis of 100 g of sample

mass of C = %C * 100 = 56.8 g

mass of H = %h 100 = 6.56g

mass of O = %O * 100 = 28.4g

mass of N = %N * 100 = 8.28g

change all masses to mol via

mol = mass/MW, where MW = molar mass of species

mass of C = 56.8/12 = 4.73

mass of H = 6.56/1 = 6.56

mass of O = 28.4/16 = 1.775

mass of N = 8.28/14 = 0.591

assume N = 1 so:

No of C =4.73 /0.591 = 8

No of H = 6.56/0.591 = 11.09

No of O = 1.775/0.591 = 3

then

For 1 mol of N

C8H11O3N1

empiical formula --> C8H11O3N

Q2.

3 isotopes

MW avg = x1*MW1+x2*MW2+x3*MW3

substitute

24.30500 = 0.79*23.99 + 0.10* 24.99 + x3 * 25.99

solve for x3

(24.30500 - 0.79*23.99 -  0.10* 24.99)/(25.99) = x3

x3 = 0.109807

x3 = 10.98 --> 11%

choose B

Q3.

mol of Xe --> mass/MW = 0.526/131.293 = 0.004

then, compound mass

m = 0.678 g, of which 0.526 g is Xe

mass of F = 0.678 -0.526 = 0.152 g is F

mol of F = mass/MW = 0.152/18.99 = 0.008

ratio

0.008/0.004 = 2

2 mol of F per 1 ml of Xe

XeF2