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vovikov84 [41]
3 years ago
6

Draw the lewis structure of ch3nc; fill in any nonbonding electrons. Calculate the formal charge on each atom other than hydroge

n.

Chemistry
1 answer:
jenyasd209 [6]3 years ago
6 0

A visual representation of covalent bonding which represents the valence shell electrons in the molecule is said to be a Lewis structure. The lines represents the shared electron pairs and dots represents the electrons that are not involved in the bonding i.e lone pairs.

Number of valence electrons in each atom:

For Carbon, C = 4

For Hydrogen, H = 1

For Nitrogen, N = 5

The Lewis structure of CH_3NC is shown in the attached image.

The formula of calculating formula charge = FC = VE - NE- \frac{BE}{2}   -(1)

where, F.C is formal charge, V.E is number of valence electrons, N.E is number of non-bonding electrons and B.E is number of bonding electrons.

Now, calculating the formal charge:

For C on left side:

FC = 4 - 0- \frac{8}{2} = 0

For N:

FC = 5 - 0- \frac{8}{2} = +1

For C on right side:

FC = 4 - 2- \frac{6}{2} = -1

The formula charge of each atom other than hydrogen is shown in the attached image.


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We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

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C. 49.48 g

H, 5.19 g

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0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

C_4H_5O_1N_2

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

C_8H_{10}O_2N_4

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Answer:

The density remains the same

Explanation:

The density remains the same because cutting the object in half will divide the mass & volume by the same amount. The density cant be changed no matter what happens to it.

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