Question

color by number
counting atoms
name:
date: 10-27-23 per: 2nd
key vocabulary

1. a subscript is a number that is written to the lower right of a symbol. the subscript tells you the

number of atoms of a particular element and only refers to the element that it is behind. if there is
no subscript present, then there is only one atom of that element. for example, in the formula
h₂o, there are 2 hydrogen atoms and only 1 oxygen atom.

1. a coefficient is a number that is placed in front of a chemical formula. you must multiply the

coefficient by the subscript to find out the number of atoms of each element. for instance, in the
formula 2h₂o, there are 4 hydrogen atoms and 2 oxygen atoms.

1. if there are parenthesis around an element or a compound, then you must multiply the subscript

behind the parenthesis by every subscript inside the parenthesis. for example, in the formula
fe(oh)₂, there is 1 iron atom, 2 oxygen atoms, and 2 hydrogen atoms.
directions: answer each question by writing the number of atoms for each element. then, circle the
total number of atoms in the chemical formula. after you have finished all questions, color the picture
on the back using the corresponding color choice.

1. h₂so₄

h =
six yellow
seven red
nine purple
s =
o =

1. ch₃oh

c =
four black
six orange
seven green
h =
o =

1. nah₂po₄

na =
four red
six black
eight yellow
h =
p =
o =

1. nh₄cl

n =
five purple
six green
seven orange
h =
cl =

1. al(oh)₃

al=
seven sky blue
five orange
four purple
o=
h=

1. fe₂(so₄)₃

fe =
seventeen yellow
ten green
fourteen orange
s =
o =

1. 3co₂ + 4h₂o

c =
fourteen red
eighteen purple
twenty - one green
h =
o =

1. al₂(so₄)₃

al = 2
fifteen green
ten red
seventeen blue
s = 2×3 = 6
o = 6×4 = 24

1. 2naoh + h₂

na = 2×1 = 2
nine green
eight red
seven blue
o = 1×1 = 1
h = 2×1 + 2 = 4

1. 4ca(hco₃)₂

ca =
twelve orange
twenty - four purple
forty - four sky blue
h =
c =
o =

1. 4li₂o

li = 4
twelve green
seven yellow
nine black
o = 8

1. 5znso₄

zn = 5
ten yellow
thirty red
fourteen purple
s = 5
o = 20

1. cu(no₃)₂

cu =
seven red
eight yellow
nine orange
n =
o =

1. 3pb(no₃)₂

pb = 3
twenty - seven blue
ten yellow
eleven green
n = 3×2 = 6
o = 3×3 = 9×2 = 18

Explanation:

Step1: Analyze \(H_2SO_4\)

For \(H\): Subscript is \(2\).
For \(S\): Subscript is \(1\).
For \(O\): Subscript is \(4\).
Total atoms \(=2 + 1+4=7\).

Step2: Analyze \(CH_3OH\)

For \(C\): Subscript is \(1\).
For \(H\): \(3 + 1=4\).
For \(O\): Subscript is \(1\).
Total atoms \(=1+4 + 1=6\).

Step3: Analyze \(NaH_2PO_4\)

For \(Na\): Subscript is \(1\).
For \(H\): Subscript is \(2\).
For \(P\): Subscript is \(1\).
For \(O\): Subscript is \(4\).
Total atoms \(=1+2 + 1+4=8\).

Step4: Analyze \(NH_4Cl\)

For \(N\): Subscript is \(1\).
For \(H\): Subscript is \(4\).
For \(Cl\): Subscript is \(1\).
Total atoms \(=1+4 + 1=6\).

Step5: Analyze \(Al(OH)_3\)

For \(Al\): Subscript is \(1\).
For \(O\): \(1\times3 = 3\).
For \(H\): \(1\times3 = 3\).
Total atoms \(=1+3+3 = 7\).

Step6: Analyze \(Fe_2(SO_4)_3\)

For \(Fe\): Subscript is \(2\).
For \(S\): \(1\times3 = 3\).
For \(O\): \(4\times3 = 12\).
Total atoms \(=2+3 + 12=17\).

Step7: Analyze \(3CO_2+4H_2O\)

For \(C\): \(1\times3 = 3\).
For \(H\): \(2\times4 = 8\).
For \(O\): \((2\times3)+(1\times4)=6 + 4=10\).
Total atoms \(=3+8 + 10=21\).

Step8: Analyze \(Al_2(SO_4)_3\)

For \(Al\): Subscript is \(2\).
For \(S\): \(1\times3 = 3\).
For \(O\): \(4\times3 = 12\).
Total atoms \(=2+3 + 12=17\).

Step9: Analyze \(2NaOH + H_2\)

For \(Na\): \(1\times2 = 2\).
For \(O\): \(1\times2 = 2\).
For \(H\): \((1\times2)+(2\times1)=2 + 2=4\).
Total atoms \(=2+2 + 4=8\).

Step10: Analyze \(4Ca(HCO_3)_2\)

For \(Ca\): \(1\times4 = 4\).
For \(H\): \(1\times2\times4 = 8\).
For \(C\): \(1\times2\times4 = 8\).
For \(O\): \(3\times2\times4 = 24\).
Total atoms \(=4+8 + 8+24=44\).

Step11: Analyze \(4Li_2O\)

For \(Li\): \(2\times4 = 8\).
For \(O\): \(1\times4 = 4\).
Total atoms \(=8+4 = 12\).

Step12: Analyze \(5ZnSO_4\)

For \(Zn\): \(1\times5 = 5\).
For \(S\): \(1\times5 = 5\).
For \(O\): \(4\times5 = 20\).
Total atoms \(=5+5 + 20=30\).

Step13: Analyze \(Cu(NO_3)_2\)

For \(Cu\): Subscript is \(1\).
For \(N\): \(1\times2 = 2\).
For \(O\): \(3\times2 = 6\).
Total atoms \(=1+2 + 6=9\).

Step14: Analyze \(3Pb(NO_3)_2\)

For \(Pb\): \(1\times3 = 3\).
For \(N\): \(1\times2\times3 = 6\).
For \(O\): \(3\times2\times3 = 18\).
Total atoms \(=3+6 + 18=27\).

Answer:

1. \(H = 2\), \(S = 1\), \(O = 4\), total \(7\) (red).
2. \(C = 1\), \(H = 4\), \(O = 1\), total \(6\) (orange).
3. \(Na = 1\), \(H = 2\), \(P = 1\), \(O = 4\), total \(8\) (yellow).
4. \(N = 1\), \(H = 4\), \(Cl = 1\), total \(6\) (green).
5. \(Al = 1\), \(O = 3\), \(H = 3\), total \(7\) (sky - blue).
6. \(Fe = 2\), \(S = 3\), \(O = 12\), total \(17\) (yellow).
7. \(C = 3\), \(H = 8\), \(O = 10\), total \(21\) (green).
8. \(Al = 2\), \(S = 3\), \(O = 12\), total \(17\) (blue).
9. \(Na = 2\), \(O = 2\), \(H = 4\), total \(8\) (red).
10. \(Ca = 4\), \(H = 8\), \(C = 8\), \(O = 24\), total \(44\) (sky - blue).
11. \(Li = 8\), \(O = 4\), total \(12\) (green).
12. \(Zn = 5\), \(S = 5\), \(O = 20\), total \(30\) (red).
13. \(Cu = 1\), \(N = 2\), \(O = 6\), total \(9\) (orange).
14. \(Pb = 3\), \(N = 6\), \(O = 18\), total \(27\) (blue).