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2 years ago

How do the percent compositions for C3H6 and C4H7 compare?

Chemistry

1 answer:

mariarad [96]2 years ago

3 0

A. They are the same

<h3>Further explanation</h3>

Given

C3H6 and C4H8

Required

The percent compositions

Solution

C₃H₆(MW = 42 g/mol)

%C = 3.12/42 x 100% = 85.71%

%H = 6.1/42 x 1005 = 14.29%

C₄H₈(MW=56 g/mol)

%C = 4.12/56 x 100% = 85.71%

%H = 8.1/56 x 100%=14.29%

So they are the same, because mol ratio of C and H in both compounds is the same, 1: 2

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For many years chloroform (CHCl3) was used as an inhalation anesthetic in spite of the fact that it is also a toxic substance th

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Hey there!:

Molar mass:

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C% = ( atomic mass C / molar mass CHCl3 ) * 100

For C :

C % = (12.01 / 119.37 ) * 100

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4 0

3 years ago

Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E

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Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

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⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

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⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

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3 years ago

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3 years ago

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