Which electron configuration represents the element carbon

. A compound contains only C, H, N, and O. It contains 37.0 % C and 42.5% O (both by mass), and there are 2 O atoms for every 1

jasenka [17]

Answer:

C₂₁H₁₅N₉O₁₈

Explanation:

Molecular formula is the ratio of atoms that are present in 1 molecule of the compound. We need to find moles of all atoms to find this ratio.

In a basis of 100, moles of C and O are:

Moles C:

37.0 * (1mol / 12g) = 3.083 moles C

Moles O:

42.5g * (1mol / 16) = 2.656 moles O

Now, to find moles of H, we need determine moles of H2O produced:

0.0310g H2O * (1mol / 18g) = 1.72x10⁻³ moles H2O * 2 = 3.44x10⁻³ moles H

These moles were produced when 0.157g of the compound react. In a basis of 100g:

3.44x10⁻³ moles H * (100g / 0.157g) = 2.194 moles H

In the same way, moles of N are:

0.0230g NH3 * (1mol / 17g) = 1.35x10⁻³ moles H2O

These moles were produced when 0.103g of the compound react. In a basis of 100g:

1.35x10⁻³ moles H * (100g / 0.103g) = 1.313 moles N

Empirical formula is (The simplest whole number ratio of atoms presents in a molecule). Dividing in the low number of moles (Moles N):

C: 3.083 moles C / 1.313 moles N = 2.3

O: 2.656 moles O / 1.313 moles N = 2.0

N: 1.313 moles N / 1.313 moles N = 1

H: 2.194 moles H / 1.313 moles N = 1.67

This ratio times 3 (To have the whole number ratio):

C: 7

O: 6

N: 3

H: 5

The empirical formula is:

C₇H₅N₃O₆

And weighs:

C: 7*(12g/mol)= 84

H: 5 * (1g/mol) = 5

N: 3 * (14g/mol) = 42

O: 6* (16g/mol) = 96

227g/mol

As the molecular mass of the compound is 681g/mol:

681 / 227 = 3

The empirical formula times 3 is the molecular formula, that is:

C₇H₅N₃O₆ × 3

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Explanation:

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It has many different uses like being used to make the tube of a vacuum cleaner and also can be used to make oil in space crafts.

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Read 2 more answers

How many liters of 1.0 M HCl do you need to neutralize 2.0 L of 3.0 M NaOH? How many liters of 1.0 M HCl do you need to neutrali

Kruka [31]

Answer:

For 1: The volume of HCl required is 6 L.

For 2: The volume of HCl required is 9 L.

For 3: The volume of sulfuric acid required is 4.5 L.

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$ ......(1)

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base

For 1:

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L$

Putting values in equation 1, we get:

$1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L$

Hence, the volume of HCl required is 6 L.

For 2:

We are given:

$n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L$

Putting values in equation 1, we get:

$1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L$

Hence, the volume of HCl required is 9 L.

For 3:

To calculate the volume of acid, we use the equation:

$N_1V_1=N_2V_2$

where,

$N_1\text{ and }V_1$ are the normality and volume of acid

$N_2\text{ and }V_2$ are the normality and volume of base

We are given:

$N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L$

Putting values in above equation, we get:

$1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L$

Hence, the volume of sulfuric acid required is 4.5 L.

8 0

3 years ago