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tiny-mole [99]
3 years ago
5

Give the mass percent of carbon in C14H19NO2?

Chemistry
2 answers:
Pachacha [2.7K]3 years ago
7 0
72.0726 is the percent for carbon
denpristay [2]3 years ago
4 0

Answer:

The mass percent of carbon in C₁₄H₁₉NO₂ is 72.1%.

Explanation:

To calculate the mass percent of carbon in C₁₄H₁₉NO₂ we have to follow two simple steps:

1st) Look for the <u>atomic weight of each element</u> in the Periodic Table and then <u>multiply each one by its subscript</u> in the molecule and finally <u>sum all</u> of them to find the molar weight of the molecule:

  • Carbon atomic weight . Carbon subscript = 12 . 14 = 168g
  • Hydrogen atomic weight . Hydrogen subscript = 1 . 19 = 19g
  • Nitrogen atomic weight . Nitrogen subscript = 14 . 1 = 14g
  • Oxygen atomic weight . Oxygen subscript = 16 . 2 = 32g

The molar weight of C₁₄H₁₉NO₂ is: 168g + 19g + 14g +32g = 233g.

2nd) Find the mass percent of carbon with a <u>Rule of three</u>:

If 233g represents the 100% of mass, the 168g of carbon in the molecule will represents the 72.1% of mass.

The Rule of three is:

233g ------------- 100%

168g ------------- = (168 . 100)/233 = 72.1% mass of carbon.

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
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Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

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At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

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\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

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