Chemistry, 29.06.2019 02:30 Niyah3406

The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. a 1.501 g1.501 g sample of ether was combusted in an oxygen rich environment to produce 3.565 g3.565 g of co2(g)co2(g) and 1.824 g1.824 g of h2o(g)h2o(g) . insert subscripts to complete the empirical formula of ether. empirical formula: cho

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Answer from: anniekwilbourne

The empirical formula of the ether will be $C_4H_{10}O$

Explanation:

The chemical equation for the combustion of ether 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=3.565g

Mass of H_2O=1.824g

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 3.565 g of carbon dioxide, $\frac{12}{44}\times 3.565=0.972g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.824 g of water, $\frac{2}{18}\times 1.824=0.202g$ of hydrogen will be contained.

Mass of oxygen in the compound = (1.501) - (0.972 + 0.202) = 0.327 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.972g}{12g/mole}=0.081moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.202g}{1g/mole}=0.202moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.327g}{16g/mole}=0.0204moles$

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 0.0204 moles.

For Carbon = $\frac{0.081}{0.0204}=3.97\approx 4$

For Hydrogen = $\frac{0.202}{0.0204}=9.9\approx 10$

For Oxygen = $\frac{0.0204}{0.0204}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is $C_4H_{10}O_1=C_4H_{10}O$

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

Answer from: Quest

super sorry but i dont know lo

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