Chemistry 1083 Name I. Determine the empirical and molecular formula of each of

Question

Chemistry 1083 Name I. Determine the empirical and molecular formula of each of the following a. Nicotine (An addictive stimulant found in nightshade family of plants like compounds. tobacco) is 74.0% carbon 8.65% hydrogen 17.35 % nitrogen. MM-162g/mole b. Cadaverine (the smell ofrancid meat): 58.77% carbon 13.81% hydrogen 27.40% nitrogen. MM-102.2g/mole ll. % mass combustion problem. 1.00g of LSD (D-lysergic acid diethy lamide) is burned and a compound containing carbon, hydrogen, nitrogen and oxygen is burned. It is determined to be 11.17% nitrogen. The combustion of this compound creates carbon dioxide and water. There are 2.81g of CO2 are formed and 0.718g of water are formed. What is the mass empirical formula of LSD?

Explanation / Answer

Q1

a)

empirical formula is the formula of least molecular coefficients

assume a basis of 162 g

then

mass of C = 74/100*162 = 119.88

mass of H = 8.65/100*162 = 14.013

mass of N = 17.35/100*162 = 28.107

change to moles

mol of C= mass/MW = 119.88/12 = 10

mol of H = mass/MW = 14.013/1 = 14

mol of N =mass/MW =  28.107/14 = 2

then

C10H14N2

empirical --> C5H7N

b)

Cadaverine

similarly, assume basis of 100 g

58.77 g of C --> mol o fC =mass/MW = 58.77/12 = 4.8975 = 5

13.81 g of H --> mol of H =mass/MW = 13.81 /1= 13.81 = 14

27.4 g of N --> mol of N = mass/MW = 27.4 /14 = 1.96 = 2

C5H14N2

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