a. silicon b. carbon c. beryllium d.chromium

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb

Bogdan [553]

Answer: The molecular formula for the menthol 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 $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

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

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

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 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen = $\frac{0.00066}{0.00066}=1$

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 $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$

4 0

2 years ago

Read 2 more answers

When 11.12 g of neon is combined in a 100 L container at 80oC with 9.59 g of argon, what is the mole fraction of argon?

PolarNik [594]

The correct answer for the question that is being presented above is this one: "0.3."

Here it is how to solve.
Molecular mass of Ar = 40
Molecular mass of Ne = 20
Number of moles of Ar = 9.59/40 = 0.239
Number of moles of Ne = 11.12/20= 0.556
Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3

3 0

3 years ago

Why are metals alloyed?

Lady_Fox [76]

Because alloys have qualities that are preferable to pure metals, almost all metals are utilised as alloys—that is, mixes of several elements. Alloying is done for a variety of purposes, the most common of which are to increase strength, corrosion resistance, or cost savings.

5 0

2 years ago

Need some help please?

VladimirAG [237]

26. B. 28. B. 20. A.

4 0

1 year ago

What changes in space result from the death of massive stars?

lana [24]

Whenever the fuel is being used up, a star explodes and the energy leakage from a star's core ceases.

Explanation:

The dying star expands in the "Red Giant," before even the inevitable collapse starts, due to nuclear reactions just outside of the core.

It becomes a white dwarf star when the star has almost the same density as the Sun. If it's much larger, a supernova explosion could take place and leave a neutron star away. However, if it is very large–at least three times the Sun's mass–the crumbling core of the star, nothing will ever stop it from crumbling. The star is imploding into a black hole, an endless gravitational loop in space.

6 0

3 years ago

￼a. silicon b. carbon c. beryllium d.chromium

a. silicon b. carbon c. beryllium d.chromium