Asample of a compound containing boron (b) and hydrogen (h) contains 5.443 g of b and 1.522 g of h. the molar mass of the compound is about 30 g. what is its molecular formula?

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

first you divide the mass of BORON by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass.

So, 5.443/11 is about 0.5. Then, the RFM of H is 1, so the number of mole is 1.522/1=1.522.

Next, you get the number of moles in order is; 0.5 and 1.522. Now we need to look at the ratio between these numbers. 0.5 is smaller so we use it as the ratio of 1.  next use 1.522/0.5 is 3.044 which has a greatest common factor of 3. so the empirical formula is BH3.

Now we are going to solve the molecular formula.

the molar mass ofthis compound is 30g, so we're going to find the RFM of the empirical formulsof BH3 first.

11+3=14.

now we see how many times 14 goes into 30. 30/14=2.14 which is about 2.

So now we need to times the subscript of the empirical formula by two.

thus, the molecular formula is B2H6.

To solve this kind of  questions, there are many steps:Know what you are calculating about, it's about the molecular formula, so you need to find out the number of moles of each elements. then use the molar mass of the whole compound to calculate the molecular formula.

1) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4) then use the molarmass of the whole compound to find the molecular formula. molar mass of the whole compound/RFM(molar mass) of the empirical formula of elements= the number you need to multiply by the subscript of the empirical formula to get the molecular formula.

please tell me if i got anything wrong;)

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.