How do you determine the formula and molar mass of silver cyanide?

Dec 26, 2016

Identify the bond and participating atoms.

Add the atomic mass of all elements present.

Explanation:

Formula

You must identify the type of bond(s) present. In this case, it is an ionic compound.

You must also identify the elements/polyatomic ions present. In this case, there is silver ($A g$), and cyanide ($C N$).
- If there was a multivalent element, you have to look at the chemical name. If there are Roman numerals, then the preceding element is multivalent with a charge of whatever the numerals indicate. E.g. $C u S {O}_{\text{4}}$ = Copper (II) sulfate -> the subscripts were reduced.

Knowing that it is an ionic compound, we know that the electrons are transferred. Thus the cation goes first ($A g$), then the anion ($C N$). The charges are $1 +$ and $1 -$, therefore we do not need to reduce.

We get $A g C N$.

Molar mass

You need a periodic table for this. The molar mass of a compound is simply the sum of all elements in the compound (including the subscripts if any; no coefficients if any).

$A g$ has a atomic mass of $107.87 \mu$
$C$ has a atomic mass of $12.01 \mu$
$N$ has a atomic mass of $14.01 \mu$

Add them all up and you get $133.89 \frac{g}{m o l}$. There is a reason why the units for atomic mass and molar mass are different. Once you learn about mols and such, you will know why, so I will leave that to your teacher.

Hope this helps :)