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3 years ago

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Which of the following describe chemical properties

1 answer:

alexandr1967 [171]3 years ago

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Chemical properties are all those properties that are shown when the chemical is a part of a reaction or is changed in any way. Unlike physical properties, they affect the composition of the chemical and the way it behaves, while physical does not change the chemical itself. Such properties are toxicity, flammability, reaction with other elements, stability, and many more.

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When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 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 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

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

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

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Hope This Helps You!
Good Luck Studying :)

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