Answer
The simplest formula of copper(II) tartrate is $CuC_{4}H_{4}O_{6}$.
The simplest formula of nitrosyl fluoroborate is $NOBF_{4}$.
Work Step by Step
If we take a sample of 100 g of copper(II) tartrate, there would be 30.03 g of Cu, 22.70 g of C, 1.91 g of H, and 45.37 g of O.
Then we have to calculate the number of moles of each element and compare them with each other.
$n (Cu) = \frac{m(Cu)}{M(Cu)} =\frac{30.03}{63.55}$ mol $= 0.473$ mol
$n (C) = \frac{m(C)}{M(C)} =\frac{22.70}{12.01}$ mol $=1.890$ mol
$ n(H) = \frac{m(H)}{M(H)} =\frac{1.91}{1.008}$ mol $=1.895$ mol
$ n(O) = \frac{m(O)}{M(O)} =\frac{42.53}{16.00}$ mol $=2.836$ mol
Now, take the smallest number of moles and use it to divide other numbers with it. Then, round that number to a whole number.
$\frac{n(Cu)}{n(Cu)}=1$
$\frac{n(C)}{n(Cu)}=4$
$\frac{n(H)}{n(Cu)}=4$
$\frac{n(O)}{n(Cu)}=6$
The simplest formula is $CuC_{4}H_{4}O_{6}$.
Follow the same procedure for nitrosyl fluoroborate.
$n (Cu) = \frac{m(N)}{M(N} =\frac{11.99}{14.01}$ mol $= 0.856$ mol
$n (Cu) = \frac{m(O)}{M(O)} =\frac{13.07}{16.00}$ mol $= 0.856$ mol
$n (Cu) = \frac{m(B)}{M(B)} =\frac{9.25}{10.81}$ mol $= 0.856$ mol
$n (Cu) = \frac{m(F)}{M(F)} =\frac{65.06}{18.998}$ mol $= 3.424$ mol
$\frac{n(N)}{n(N)}=1$
$\frac{n(O)}{n(N)}=1$
$\frac{n(B)}{n(N)}=1$
$\frac{n(F)}{n(N)}=4$
The simplest formula is $NOBF_{4}$.
