Question

Hat mass of ethane, c2h6, contains the same number of molecules as 3.00 g of trichlorofluoromethane, ccl3f?

Answer 1

The mass of ${\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}$ is 0.654 g

Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.

Step1: Initially we have to calculate the moles of trichlorofluoromethane $\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$

The formula to calculate the moles of trichlorofluoromethane $\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$ is as follows:

${\text{Moles of CC}}{{\text{l}}_{\text{3}}}{\text{F}}=\frac{{{\text{Given mass of CC}}{{\text{l}}_{\text{3}}}{\text{F}}}}{{{\text{Molar mass of CC}}{{\text{l}}_{\text{3}}}{\text{F}}}}$                                  …… (1)

The given mass of $\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$ is 3 g.

The molar mass of $\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$ is 137.34 g/mol.

Substitute these values in equation (1)

$\align{{\text{Moles of CC}}{{\text{l}}_{\text{3}}}{\text{F}}{\bf&{=}}\left({3\;{\text{g}}}\right)\left({\frac{{{\text{1}}\;{\text{mol}}}}{{{\text{137}}{\text{.34}}\;{\text{g}}}}}\right)\cr&=0.02184\;{\text{mol}}\cr}$

The number of moles of trichlorofluoromethane $\left({{\text{CC}}{{\text{l}}_{\text{3}}}{\text{F}}}\right)$ is equal to the number of moles of ethane $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ and that is 0.02184 moles.

Step2: Calculate the mass of ethane $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ from the moles of ethane $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ .

The formula to calculate the moles of ethane $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ is as follows:

${\text{Moles of C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}=\frac{{{\text{Given mass of C}}{& _{\text{2}}}{{\text{H}}_{\text{6}}}}}{{{\text{Molar mass of C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}}$                              …… (2)

The moles of $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ is 0.02184 mol.

The molar mass of $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ is 30 g/mol.

Substitute these values in equation (2)

$0.02184\;{\text{mol}}{\bf{=}}\left({{\text{Given mass of C}}{ & _{\text{2}}}{{\text{H}}_{\text{6}}}}\right)\left({\frac{{{\text{1}}\;{\text{mol}}}}{{{\text{137}}{\text{.34}}\;{\text{g}}}}}\right)$

(3)

Rearrange the equation (3) to calculate the mass of  $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$

$\align{{\text{Given mass of }}{{\text{C}}_2}{{\text{H}}_6}&=\frac{{0.02184\;{\text{mol}}}}{{{\text{137}}{\text{.34}}\;{\text{g/mol}}}}\cr&=0.654\;{\text{g}}\cr}$

Hence, the mass of $\left({{\text{C}}{&_{\text{2}}}{{\text{H}}_{\text{6}}}}\right)$ is 0.654 g.

Learn more:

1. Bond energy of H-H bond in the given reaction: brainly.com/question/7213980

2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Mole concept

Keywords: stoichiometry, C2H6, CCL3F, moles, mass, molar mass, 0.02184 mol., 30 g/mol, 3 g, 137.34 g/mol and 0.654 g.

Answer 2

1. Calculate the molar masses of each compound:

$C_2H_6 = 12\cdot 2 + 1\cdot 6 = 30\ g/mol$

$CCl_3F = 12\cdot 1 + 35.5\cdot 3 + 19\cdot 1 = 137.5\ g/mol$

2. Both masses are referred to one mole of the substances. We can do a ratio between them and the 3 g of $CCl_3F$:

$\frac{30\ g\ C_2H_6}{137.5\ g\ CCl_3F} = \frac{x}{3\ g\ CCl_3F}\ \to\ x = \bf 0.654\ g\ C_2H_6$