Question

question # | total # of atoms

1. h₂so₄

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

1. ch₃oh

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

1. nah₂po₄

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

1. nh₄cl

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

1. al(oh)₃

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

1. fe₂(so₄)₃

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

1. 3co₂ + 4h₂o

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

Explanation:

Question 1: $\boldsymbol{H_2SO_4}$

Step1: Count H atoms

In $H_2SO_4$, the subscript of H is 2, so $H = 2$.

Step2: Count S atoms

The subscript of S is 1, so $S = 1$.

Step3: Count O atoms

The subscript of O is 4, so $O = 4$.

Step4: Total atoms

Sum: $2 + 1 + 4 = 7$. So the total is seven [red].

Question 2: $\boldsymbol{CH_3OH}$

Step1: Count C atoms

In $CH_3OH$, $C = 1$.

Step2: Count H atoms

H subscripts: 3 (from $CH_3$) + 1 (from $OH$) = 4, so $H = 4$.

Step3: Count O atoms

$O = 1$.

Step4: Total atoms

Sum: $1 + 4 + 1 = 6$. So the total is six [orange].

Question 3: $\boldsymbol{NaH_2PO_4}$

Step1: Count Na atoms

$Na = 1$.

Step2: Count H atoms

$H = 2$.

Step3: Count P atoms

$P = 1$.

Step4: Count O atoms

$O = 4$.

Step5: Total atoms

Sum: $1 + 2 + 1 + 4 = 8$. So the total is eight [yellow].

Question 4: $\boldsymbol{NH_4Cl}$

Answer:

s:

1. $H_2SO_4$: seven [red]
2. $CH_3OH$: six [orange]
3. $NaH_2PO_4$: eight [yellow]
4. $NH_4Cl$: six [green]
5. $Al(OH)_3$: seven [sky blue]
6. $Fe_2(SO_4)_3$: seventeen [yellow]
7. $3CO_2 + 4H_2O$: twenty - one [green]