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Verdich [7]
3 years ago
7

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.
Chemistry
1 answer:
andrew-mc [135]3 years ago
7 0

Answer: 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

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A. Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculat
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Answer:

A)Mass of  gallium plated out is 0.3440 grams

B) For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

Explanation:

To calculate the total charge, we use the equation:

C=I\times t

where,

C = Charge

I = Current in time t (seconds)

To calculate the moles of electrons, we use the equation:

\text{Moles of electrons}=\frac{C}{F}

where,

F = Faraday's constant = 96500

A) The equation for the deposition of Ga(s) from Ga(III) solution follows:

Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)

I = 0.790 A, t = 30.0 min = 1800 seconds

C=I\times t

C=0.790 A\times 1800 s=1422 C

Moles of electron transferred:

=\frac{1422 C}{96500 F}=0.01474 mol

Now, to calculate the moles of gallium, we use the equation:

\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}

n = number of electrons transferred = 3

\text{Moles of Gallium}=\frac{0.01474 mol}{3}=0.004913 mol

Mass of 0.004913 moles of gallium = 0.004913 mol × 70 g/mol=0.3440 g

B) The equation for the deposition of Sn(s) from Sn(II) solution follows:

Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)

Moles of tin = \frac{8.70 g}{119 g/mol}=0.07311 mol

n = number of electrons transferred = 2

\text{Moles of tin}=\frac{\text{Moles of electrons}}{n}

Moles of electron =  n\times \text{Moles of tin}

=2\times 0.07311 mol=0.14622 mol

Charge transferred during time t :

\text{Moles of electrons}=\frac{C}{F}

C=96500 F\times 0.14622 mol=14,110.23 C

Current applied for t time = I = 5.79 A

t=\frac{C}{I}=\frac{14,110.23 C}{5.79 A}=2,437 s=0.67 hrs

For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.

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The answer is critical mass with no doub t
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