Mixture composition

[BeYeu05]

Question: How much of each compound is needed to make 0.5 g of 80%/20% (mole %) mixture of cinnamic acid and urea acid?

 

So far, I have the molar mass of individual compound:

cinnamic acid =148.16 g

urea = 60.062 g

 

i tried to do 80% of 0.5 g which gave 0.4 g of cinnamic acid, but i am sure that is wrong because the % is not mass/weight % but rather MOLE%, which i do not know how to do.

 

please help?

[Primarygun]

I think I know what you mean .

For every amount of the mixture, we have the ratio given/required.

Let's handle one mole first,

Mole of cinnamic acid :mole of urea acid=4:1

so, in a mixture of 1 mole of urea acid,

the mass=(148.16 x 4 g+60.062 g )=653g

When 0.5g is divided by it,

we get the no. of mole of urea required.

Then you can finish it yourself/

[Primarygun]

Or a faster method,

0.8x148.16 g:0.2x60.062 g will be the ratio of each compound in the 0.5 g