All categories

3 years ago

230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen

Chemistry

1 answer:

natulia [17]3 years ago

6 0

Answer:

15.3 %

Explanation:

Step 1: Given data

Mass of the sample (ms): 230 g

Mass of carbon (mC); 136.6 g

Mass of hydrogen (mH): 26.4 g

Mass of nitrogen (mN): 31.8 g

Step 2: Calculate the mass of oxygen (mO)

The mass of the sample is equal to the sum of the masses of all the elements.

ms = mC + mH + mN + mO

mO = ms - mC - mH - mN

mO = 230 g - 136.6 g - 26.4 g - 31.8 g

mO = 35.2 g

Step 3: Calculate the mass percent of oxygen

%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %

You might be interested in

What is name of each shape

dimulka [17.4K]

Answer:

I am so sorry I do not know this

Explanation:

4 0

2 years ago

The factor that affects how easy and electron can be removed from an Atom is the

zheka24 [161]

The ionization energy

3 0

3 years ago

Hi, can someone help me with chemistry

Dominik [7]

Idk look it up on another website

6 0

2 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

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 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

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

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

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

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

When can all the coefficients in a balanced equation represent volume ratios? when reactants and products are solids at stp when

Alexxx [7]

Answer: when reactants and products are gases at STP.

Justification:

1) STP stands for standard temperature (0°) and pressure (1 atm).

2) According to the kinetic molecular theory of the gases, and as per Avogadro's principle, equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

3) Since the coefficients in a balanced chemical equation represent number of moles, when reactants and products are gases at the same temperature and pressure, the mole ratios are the same that the volume ratios, and then the coefficients of the chemical equation represent the volume ratios.

8 0

3 years ago

Read 2 more answers