All categories

3 years ago

Determine the empirical formula of kevlar, used in making bullet proof vests, is 70.6% c, 4.2% h, 11.8% n and 13.4% o.

Chemistry

1 answer:

Likurg_2 [28]3 years ago

3 0

Assuming a basis of 100 g of Kevlar, therefore the masses are:

C = 70.6 g

H = 4.2 g

N = 11.8 g

O = 13.4 g

We know the molar masses are:

C = 12.01, H = 1.008, N = 14.01, O = 16.00

Calculating for the moles per element:

C = 70.6 / 12.01 = 5.88 mol

H = 4.2 / 1.008 = 4.17 mol

N = 11.8 / 14.01 = 0.84 mol

O = 13.4 / 16 = 0.84 mol

To get the empirical formula, the next step is to divide everything by the smallest moles = 0.84. So:

C = 7

H = 5

N = 1

O = 1

Hence the empirical formula is:

C7H5NO

You might be interested in

Name the noble gas atom that has the same electron configuration as each ion in the compound rubidium oxide

Zepler [3.9K]

In rubidium oxide - Rb₂O , the ions are Rb⁺ and O²⁻
Rb is a group one element with one valence electron. To become stable it loses its outer electron to gain a complete outer shell.
Electronic configuration of Rb is - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
Once it loses its valence electron the configuration is;
- 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶
The noble gas with this configuration is Krypton - Kr

Oxygen electron configuration is 1s² 2s² 2p⁴
Once it gains 2 electrons the configuration is - 1s² 2s² 3p⁶
The noble gas with this configuration is Neon - Ne

3 0

4 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen 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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

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 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

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

Formula used :

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

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

Help Please and answer correctly otherwise, I can report the answer and your account so make sure the answer is correct BUT PLEA

Alina [70]

Answer:

Explanation:

8 0

3 years ago

Describe the contributions made by the following scientists to the development of the periodic table.

Basile [38]

Answer:

Explanation:

dimitri Mendeleev......

6 0

3 years ago

Explain step by step, please!

konstantin123 [22]

Answer:

10. Stacey

11. 7.86 g/cm^3 (3dp)

OR 786 kg/m^3 (SI units)

Explanation:

10. The correct exact answer for 10 is 26.169 (6.71x3.9). While Sam was correct in the precision of the number (2dp just like the given measurements), he was incorrect in the number itself. While Stacy's number was not as precise as Sam's (0dp), it is correctly rounded to the nearest whole number, therefore Stacy is correct.

11. The formula for density is mass/volume. The mass here is 264g and the volume is 33.6ml, therefore the density is 7.86 (3 sig figs because given values had 3) g/cm^3, because the mass was given in grams and 1ml = 1cm^3. Converted to SI units it is 786 kg/m^3.

Hope this helped!

4 0

3 years ago