Question

1. what is the name of the compound fe(no3)3?
2. calculate the molar mass of the compound ca3(po4)2
3. what is the formula of the compound titanium (ii) nitride
4. for the compound (nh4)2co3, was is the percent composition for each element?
5. if you have 15 grams of cas, how many moles and molecules do you have?
6. what is the mass, in grams, of 2.85 x 10^22 molecules of hno3?
7. a compound contains 1.271 grams of aluminum and 5.009 grams of chlorine. what is its empirical formula?
8. what is the name of the compound n2o4?
9. how many molecules are in 48.5 grams of nabr?

Explanation:

Step1: Name of Fe(NO₃)₃

Fe is iron, NO₃⁻ is nitrate. Iron has a +3 charge here, so it's iron(III) nitrate.

Step2: Molar mass of Ca₃(PO₄)₂

Atomic masses: Ca = 40.08 g/mol, P = 30.97 g/mol, O = 16.00 g/mol.
$M = 3\times40.08+2\times(30.97 + 4\times16.00)$
$M = 120.24+2\times(30.97 + 64.00)$
$M = 120.24+2\times94.97$
$M = 120.24 + 189.94=310.18$ g/mol

Step3: Formula of titanium(II) nitride

Titanium(II) has a +2 charge (Ti²⁺), nitride is N³⁻. Using criss - cross method, the formula is Ti₃N₂.

Step4: Percent composition of (NH₄)₂CO₃

Atomic masses: N = 14.01 g/mol, H = 1.01 g/mol, C = 12.01 g/mol, O = 16.00 g/mol.
Molar mass of (NH₄)₂CO₃: $M=(2\times14.01)+(8\times1.01)+12.01+(3\times16.00)=96.09$ g/mol
Percent of N: $\frac{2\times14.01}{96.09}\times100\%\approx29.16\%$
Percent of H: $\frac{8\times1.01}{96.09}\times100\%\approx8.43\%$
Percent of C: $\frac{12.01}{96.09}\times100\%\approx12.50\%$
Percent of O: $\frac{3\times16.00}{96.09}\times100\%\approx50.00\%$

Step5: Moles and molecules of 15 g of CaS

Molar mass of CaS: $M = 40.08+32.07 = 72.15$ g/mol
Moles: $n=\frac{m}{M}=\frac{15}{72.15}\approx0.208$ mol
Molecules: $N = n\times N_A=0.208\times6.022\times10^{23}\approx1.25\times10^{23}$ molecules

Step6: Mass of 2.85×10²² molecules of HNO₃

Molar mass of HNO₃: $M = 1.01+14.01+(3\times16.00)=63.02$ g/mol
Moles: $n=\frac{N}{N_A}=\frac{2.85\times10^{22}}{6.022\times10^{23}}\approx0.0473$ mol
Mass: $m = n\times M=0.0473\times63.02\approx2.98$ g

Step7: Empirical formula of a compound with Al and Cl

Moles of Al: $n_{Al}=\frac{1.271}{26.98}\approx0.0471$ mol
Moles of Cl: $n_{Cl}=\frac{5.009}{35.45}\approx0.141$ mol
Ratio of Al:Cl = $\frac{0.0471}{0.0471}:\frac{0.141}{0.0471}\approx1:3$
Empirical formula is AlCl₃.

Step8: Name of N₂O₄

It is dinitrogen tetroxide.

Step9: Molecules in 48.5 g of NaBr

Molar mass of NaBr: $M = 22.99+79.90 = 102.89$ g/mol
Moles: $n=\frac{m}{M}=\frac{48.5}{102.89}\approx0.471$ mol
Molecules: $N = n\times N_A=0.471\times6.022\times10^{23}\approx2.84\times10^{23}$ molecules

Answer:

1. Iron(III) nitrate
2. 310.18 g/mol
3. Ti₃N₂
4. N: 29.16%, H: 8.43%, C: 12.50%, O: 50.00%
5. Moles: 0.208 mol, Molecules: $1.25\times10^{23}$ molecules
6. 2.98 g
7. AlCl₃
8. Dinitrogen tetroxide
9. $2.84\times10^{23}$ molecules