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2 years ago

Please help!! I need this by tomorrow

Chemistry

1 answer:

notka56 [123]2 years ago

6 0

Answer:

Explanation:

First, set AL2(so4)3 to 1 and use the law of conservation of mass to solve the problem.

and At first, there were only 23 atoms in the system. After the reaction, the number of atoms remained unchanged.

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Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H

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Answer : The molecular formula of a caffeine is, $C_8H_{10}N_4O_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles$

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

For C = $\frac{4.12}{1.03}=4$

For H = $\frac{5.19}{1.03}=5.03\approx 5$

For N = $\frac{2.06}{1.03}=2$

For O = $\frac{1.03}{1.03}=1$

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_4H_5N_2O_1=C_4H_5N_2O$

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{194.19}{97}=2$

Molecular formula = $(C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2$

Therefore, the molecular of the caffeine is, $C_8H_{10}N_4O_2$

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