1.When a 0.528 g sample of a compound containing only carbon, hydrogen, and oxyg

Question

1.When a 0.528 g sample of a compound containing only carbon, hydrogen, and oxygen is burned, 0.306 g of water and 0.815 g of carbon dioxide are formed. What is the empirical formula for this compound?

2. If 20.0mL of 0.300 M solution of NaOH is require to neutralize 30.0mL of a sulfuric acid solution, What is the molarity of the H2SO4?

3. A 25.0 gram piece of metal is heated to 98.0 celsius. The metal is then placed into calorimeter containing 85.0mL of water at 35 celsius. The temperature of the calorimeter when the metal is added goes up to 77 celsius. What is the specific heat of the metal?

Explanation / Answer

(1)find grams, using molar masses:

0.815 g CO2 @ 12.0 g C / 44.0 g CO2 = 0.222 grams C
0.306 g H2O @ 2.02 g H / 18 g H2O = 0.034 grams H
0.528 g sample - 0.222 g C - 0.034 g H = 0.272 grams O

find moles, using molar masses:
0.222 grams C / 12.0 g/mol C = 0.0185 moles C
0.034 grams H / 1.01 g/mol H = 0.033 moles H
0.272 grams O / 16.0 g/mol O = 0.017 moles O

find ratios:
0.0185 moles C / 0.017 = 1 moles C
0.033 moles H / 0.017= 2 moles H
0.017 moles O / 0.017 = 1 mole O

double the ratio:
the ratio is C1 H2 O 1

gives your answer is C2 H4 O2

(2)

H2SO4 + 2 NaOH --> Na2SO4 + 2H2O
2 millimoles of NaOH are required to neutralize 1 millimole of H2SO4

20.0 mL x 0.30 M = 6 millimoles of NaOH
30.0 mL x ? M = 3 millimoles of H2SO4
? M = 0.10 M H2SO4

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.