30. The natural abundance of 35CI is 75.77%. These means that in 100 g of Cl the

Question

30. The natural abundance of 35CI is 75.77%. These means that in 100 g of Cl there are 75.77 g of 35CI. A. one B. two D. nine C. three E. infinite 29. An unknown carbon, hydrogen, and oxygen containing compound was combusted. A 4.50 gram sample of the unknown was combusted yielding 6.60 g of CO2 and 2/0 g of H2O. What is the empirical formula? A. CHO B. CH2O C. C2H3O D. C2H6O E. none 31. What is the natural abundance of 2H-2H? 2H is 0.015% abundant. B. 0.030% D. 0.0000023 0/0 A. 0.015% C. 0.00023%

Explanation / Answer

Q30 Solution

Since the percent abundance if the 35Cl is 75.77 %

Therefore in 100 g sample mass of the 35Cl is 75.77 g

1 Cl = 35

Therefore 75.77 / 35 = 2

Therefore it have Two 35Cl Therefore option B is the correct

Q29 Solution

Mass of sample =4.50 g

Mass of CO2 =6.60 g

Mass of H2O = 2.70 g

Empirical formula

Lets first calculate the mass of the C and H using the mass percent of the C and H in CO2 and H2O

6.60 g CO2* 27.3% / 100 % = 1.80 g C

2.70 g H2O * 11.2% / 100 % = 0.3024 g H

Mass of oxygen = 4.50 g – ( mass of C + mass of H)

                             = 4.50 g – (1.80 g + 0.3024g)

                             = 2.4 g

Now lets calculate moles of the each element

Moles of C = 1.80 g / 12.01 g per mol = 0.15 mol C

Moles of H = 0.3024 g / 1.0079 g per mol = 0.30 mol H

Moles of O = 2.4 g / 16 g per mol = 0.15 mol O

Now lets calculate the ratio of the each element by dividing each moles by smallest mole value

C= 0.15 /0.15 = 1

H=0.30/0.15 = 2

O = 0.15/0.15 = 1

Empirical formula CnHnOn = CH2O

Therefore answer is option B

Q31 solution :-

Since natural abundance of the 2H is 0.015%

Therefore in the H2 molecule also the natural abundance of the 2H is 0.015 %

Hence option A

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