What is the empirical formula for c3n12

The emoirical formula for the given compound is :

. It is simply written as CN4

Explanation:

Molecular formula : It  is the symbolic representation of chemical composition of different elements present in the molecule.

For example : Benzene is C6H6

It is the true composition of the elements.

Empirical Formula :It is the simple whole number ratio of the elements present in the given molecule.

For example :

The empirical formula of  Benzene is:

= CH

Steps to reach at the empirical formula :

In C3N12

3 is the common factor

divide each term by 3

C3 on dividing by 3 = C1

N12 on dividing by 3 = N4

Hence the empirical formula of the compound is CN4

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer