17 (3 pts) Assuming that it could exist, what would be the mass percent of carbo

Question

17 (3 pts) Assuming that it could exist, what would be the mass percent of carbon in carbon trioxide (FM 60.01)? %C = 18 (6 pts) A certain compound contains only C and H and has a molar mass of about 85 g/mol. A sample of the compound burns in a combustion apparatus yielding 5.12 g CO2 and 2.08 g H2O. What is its molecular formula? molecular formula= 19. (5 pts) A compound of carbon and hydrogen contains 44.35 g carbon and 11.32 g hydrogen. How many grains of carbon are in 48.8 g of this compound? mass C = 20 (2 pts) Using an example of each, explain how a strong mid differs from a weak acid.

Explanation / Answer

17. 1 molecule of carbon trioxide has 1 carbon in it

Mass of carbon = 12.01 g

mass of carbn trioxide = 60.01 g

So, mass percentage of carbon = 12.01/60.01 = 20.01%

18. Find mass percentage,

C = 5.12 g = 5.12 x 12/44 = 1.40

H = 2.08 g = 2.08 x 2/18 = 0.23

Total mass = C + H = 1.40 + 0.23 = 1.63 g

mass % of C = 1.40/1.63 = 86%

mass % of H = 0.23/1.63 = 14%

finding empirical formula,

C = 86/12 = 7.17

H = 14/1 = 14

Divide by smallest number

C = 7.17/7.17 = 1

H = 14/7.17 = 2

Emirical formula = CH2

Empirical formula weight = 12 + 2 = 14 g

ratio = molar mass/empirical formula weight = 85/14 = 6

Molecular formula = C6H12

19. Compound has,

C = 44.35 g

H = 11.32 g

Total weight = C + H = 44.35 + 11.32 = 55.67 g

% of carbon in sample = 44.35 x 100/55.67 = 79.70%

So, grams of carbon in 48.8 g sample wouls be = 79.70x 48.8/100 = 38.88 g

20. Strong acid such as HCl dissociates completely when in solution,

HCl(aq) ----> H+(aq) + Cl-(aq)

the pH of solution is thus, = -log[conc. of HCl]

A weak acid such as acetic acid, dissociates only partially in solution to produce H+ ion,

CH3COOH(aq) <===> CH3COO-(aq) + H+(aq)

the pH of solution would be dependent on how much acid is dissociated at equilibrium

Empirical formula weight = 12 + 5 x 1 = 17 g

Ratio = molecular formula weight/empirical formula weight

         = 85/17 = 5

So molecular formula becomes, C5H25

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