Question

a certain alkaloid has 70.8% carbon, 6.2%hydrogen, 4.1%nitrogen and the rest oxygen.What is its empirical formula?

Answer 1

Hello!

a certain alkaloid has 70.8% carbon, 6.2%hydrogen, 4.1%nitrogen and the rest oxygen.What is its empirical formula?

data:

Carbon (C) ≈ 12 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Nitrogen (N) ≈ 14 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 70.8 % → 70.8 g

H: 6.2 % → 6.2 g

N: 4.1 % → 4.1 g

O:  (100% - 70.8% - 6.2% - 4.1%) = 18.9 % → 18.9 g

The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

$C: \dfrac{70.8\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 5.9\:mol$

$H: \dfrac{6.2\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 6.2\:mol$

$N: \dfrac{4.1\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.29\:mol$

$O: \dfrac{18.9\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} \approx 1.18\:mol$

We realize that the values ​​found above, some are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let's see:

$C: \dfrac{5.9}{0.29}\to\:\:\boxed{C = 20.3}$

$H: \dfrac{6.2}{0.29}\to\:\:\boxed{H \approx 21.3}$

$N: \dfrac{0.29}{0.29}\to\:\:\boxed{N = 1}$

$O: \dfrac{1.18}{0.29}\to\:\:\boxed{O \approx 4}$

= 20 : 21 : 1 : 4  ← whole number of atomic radio  

C ≈ 20

H ≈ 21

N = 1

O ≈ 4

Thus, the minimum or Empirical Formula (E.F) found for the compound will be:

$\boxed{\boxed{C_{20}H_{21}NO_4}}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark$

Answer:

C20H21NO4

__________________________  

I Hope this helps, greetings ... Dexteright02! =)

Answer 2

Answer: The empirical formula for this alkaloid is $C_{20}H_{21}N_1O_4$

Explanation:

To write the empirial formula, we need to follow some steps:

• Step 1: Converting percentages to masses

If percentage are given then we are taking total mass is 100 grams.  Thus, the mass of each element is equal to the percentage given.

Mass of C = 70.8 g

Mass of H = 6.2 g

Mass of N = 4.1 g

Mass of O = 18.9 g

• Step 2: Converting given masses into moles.

$\text{Moles of C}=\frac{\text{Given mass of C}}{\text{Molar mass of C}}= \frac{70.8g}{12g/mole}=5.9moles$

$\text{Moles of H}=\frac{\text{Given mass of H}}{\text{Molar mass of H}}= \frac{6.2g}{1g/mole}=6.2moles$

$\text{Moles of N}=\frac{\text{Given mass of N}}{\text{Molar mass of N}}= \frac{4.1g}{14g/mole}=0.292mole$

$\text{Moles of O}=\frac{\text{Given mass of O}}{\text{Molar mass of O}}= \frac{18.9g}{16g/mole}=1.181moles$

• Step 3: Converting the moles into mole ratios

For calculating the mole ratio, we divide each value of moles by the smallest number of moles calculated which is 0.292 moles.

For C = $\frac{5.9}{0.292}=20.20\approx 20$

For H = $\frac{6.1}{0.292}=21.23\approx 21$

For N= $\frac{0.292}{0.292}=1$

For O =$\frac{1.181}{0.292}=4.04\approx 4$

• Step 4: Now writing the mole fraction as the subscripts of the elements.

The ratio of C : H : N : O= 20 : 21 : 1 : 4

Hence the empirical formula is $C_{20}H_{21}N_1O_4$