Chemistry, 17.06.2020 21:57 fffcc7665

A 3.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 192./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 4.13g water 1.13g Use this information to find the molecular formula of X.

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Answer from: marcgtz511p3pln7

The molecular formula for the given organic compound X is $C_6H_{8}O_7$

Explanation:

We are given:

Mass of CO_2=4.13g

Mass of H_2O=1.13g

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 4.13 g of carbon dioxide, = $\frac{12}{44}\times 4.13=1.13g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.13 g of water, $\frac{2}{18}\times 1.13=0.125g$ of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles$

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

For Carbon = $\frac{0.094}{0.094}=1$

For Hydrogen = $\frac{0.125}{0.094}=1.33$

For Oxygen = $\frac{0.109}{0.094}=1.16$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is C_6H_8O_7

Empirical mass = $6\times 12+8\times 1+7\times 16=192g$

The equation used to calculate the valency is :

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

Putting values in above equation, we get:

$n=\frac{192g/mol}{192g/mol}=1$

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

$C_6H_8O_7\times 1=C_6H_{8}O_7$

Thus molecular formula for the given organic compound X is $C_6H_{8}O_7$

Answer from: Quest

n2(g) + 3h2-> 2nh3(g) this is the balanced equation

note the mole ratio between n2, h2 and nh3. it is 1 : 3 : 2 this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note: the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

Answer from: Quest

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