A sample of a compound contains 60.0 g C and 5.05 g H. It’s molar mass is 78.12 g/mol. What is the compound’s molecular formula?
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Answer:
"0.053457 M" of sulfuric acid.
Explanation:
The given values are:
= 10 mL solution
= 12.20 mL
= 22.20 mL
then,
M 0.103 M of NaOH,
= experiment will not be affected
= 10.38 mL
Now,
⇒ mol of NAOH = MV
=
=
Whether Sulfuric acid, then
⇒
⇒
⇒
Before any dilution:
⇒
(Sulfuric acid)
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34
= 0.059 g
percent morphine =
=
= 8.6 %
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34
= 0.059 g
percent morphine =
=