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2 years ago

What is the molecular formula of an empirical N2H3 if n=2? SHOW WORK!

2 answers:

7 0

A compound's empirical formula tells you what the smallest whole number ratio between the atoms that make up that compound is.

I think of the empirical formula as a building block for molecules. A compound's molecular formula will depend on how many building blocks are needed to build a molecule of a given substance.

In your example, you know that the empirical formula of the compound is

This tells you that the minimum ratio between nitrogen atoms and oxygen atoms is

. In other words, the molecular formula will be a multiple of this building block.

molecular formula

empirical formula

, where

- the number of building blocks needed to build the molecular formula.

You also know that the molar mass of your compound is

92 g/mol

This means that the molar mass of all the atoms that make up that molecule must add up to give

92 g/mol

So, how would you determine how many building blocks you need? Well, start by figuring out the molar mass of one building block, i.e. the molar mass of the empirical formula.

Since it contains one nitrogen atom and two oxygen atoms, you will get

14.0067 g/mol

15.9994 g/mol

46.0055 g/mol

So, if one building block has a molar mass of

46.0055 g/mol

, how many would you need to get the molecule?

46.0055 g/mol

⋅

92 g/mol

g/mol

46.0055

g/mol

1.99976

≈

This means that the compound's molecular formula, which lists all the atoms that make up a molecule, will be

(

)

→

dinitrogen tetroxide

Papessa [141]2 years ago

7 0

Answer:

The empirical formula for a compound is The molar mass of the compound is 93 grams/mole. n=2 is 0

Explanation:

is what I got

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2 years ago

When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.