Chemistry, 19.08.2019 18:20, deniseelyons15

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 0.1005 −mg sample of menthol is combusted, producing 0.2829 mg of co2 and 0.1159 mg of h2o. what is the empirical formula for menthol? express your answer as a chemical formula.

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answered: asdfghj3513

The empirical formula for the given compound is $C_{10}H_{20}O$

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=0.2829mg=0.0002829g

Mass of H_2O=0.1159mg=0.0001159g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 0.0002829 g of carbon dioxide, $\frac{12}{44}\times 0.0002829=0.00007715g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.0001159 g of water, $\frac{2}{18}\times 0.0001159=0.00001288g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.0001005) - (0.00007715 + 0.00001288) = 0.00001047 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00007715g}{12g/mole}=6.429\times 10^{-6}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00001288g}{1g/mole}=0.01288\times 10^{-3}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.00001047g}{16g/mole}=6.544\times 10^{-7}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $6.544\times 10^{-7}moles$ .

For Carbon = $\frac{6.429\times 10^{-6}}{6.544\times 10^{-7}}=9.82\approx 10$

For Hydrogen = $\frac{0.01288\times 10^{-3}}{6.544\times 10^{-7}}=19.68\approx 20$

For Oxygen = $\frac{6.544\times 10^{-7}}{6.544\times 10^{-7}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

answered: Guest

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answered: Guest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.