Chemistry, 15.07.2020 03:01 shoreelinee1337

A 0.3832-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7120 g of CO2 and 0.1458 g of H2O. What is the empirical formula of the compound

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Answer from: 10040813

C_3H_3O_2

Explanation:

Hello,

In this case, since the law of conservation of mass allows us to notice that the mass of carbon in the burned compound is also present in the resulting carbon dioxide, we can compute such moles as shown below:

$n_C=0.7120gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.0162molC$

Next, hydrogen in the sample is present at the products in the water only, and in one mole of water, two moles of hydrogen are present, thereby:

$n_H=0.1458gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=0.0162molH$

Nevertheless, the amount of oxygen in the sample must be computed by subtracting both the mass of carbon and hydrogen from the previously computed moles:

$m_O=0.3832g-0.0162molC*\frac{12gC}{1molC} -0.0162molH*\frac{1gH}{1molH} =0.1726gO$

And the moles:

$n_O=0.1726gO*\frac{1molO}{16gO} =0.0108molO$

Next, we compute the mole ratios with respect to the element having the smallest number of moles (oxygen) to obtain the subscripts in the empirical formula:

$C=\frac{0.0162}{0.0108}=1.5\\ \\H=\frac{0.0162}{0.0108}=1.5\\\\O=\frac{0.0108}{0.0108}=1$

Then, we have:

$C_{1.5}H_{1.5}O_1$

Finally, by multiplying by two, we obtain the smallest whole subscripts:

C_3H_3O_2

Best regards.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

they it depends on if one believes about radioactivity, they make a geiger counter beep but maybe it was going to beep anyway electronics do wierd things all the time or it was programmed to beep near certain metals.

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A 0.3832-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxy...

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