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3 years ago

Cuales son los reactivos y los productos del enlace peptidico?

Chemistry

1 answer:

Hoochie [10]3 years ago

4 0

Answer:

this isnt english

Explanation:

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WORTH 20 POINTS Confirmation bias is where a scientist may change the data to support a certain bias.

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3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

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Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

6 0

3 years ago