Question

counting atoms worksheet
directions for each problem:

1. recognize and record individual elements
2. recognize the subscripts as the number of atoms for an element
3. recognize the coefficients as the number of atoms for all elements in the compound.
4. total all atoms for an element.

examples:
a) naoh
na - 1
o - 1
3 total atoms
h - 1
b) 5znso₄
zn - (5x1 = 5)
s - (5 x 1 = 5)
30 total atoms
o - 20 (5 x 4 = 20)

1. h₂o
2. mgcl₂
3. li₂so₄
4. nac₂h₃o₂
5. nh₄cl
6. hbr
7. 4hno₃
8. 4li₂o
9. 3h₂o
10. 3al₂o₃
11. 2ch₃oh
12. 2h₃nch₃br

Explanation:

Step1: Analyze $H_2O$

For $H_2O$, hydrogen ($H$) has sub - script 2, so number of $H$ atoms is 2. Oxygen ($O$) has sub - script 1 (implied when no number is written), so number of $O$ atoms is 1. Total atoms = 2 + 1=3.

Step2: Analyze $MgCl_2$

For $MgCl_2$, magnesium ($Mg$) has 1 atom (sub - script 1 implied). Chlorine ($Cl$) has sub - script 2, so number of $Cl$ atoms is 2. Total atoms = 1+2 = 3.

Step3: Analyze $Li_2SO_4$

For $Li_2SO_4$, lithium ($Li$) has sub - script 2, so number of $Li$ atoms is 2. Sulfur ($S$) has 1 atom (sub - script 1 implied). Oxygen ($O$) has sub - script 4, so number of $O$ atoms is 4. Total atoms = 2 + 1+4=7.

Step4: Analyze $NaC_2H_3O_2$

For $NaC_2H_3O_2$, sodium ($Na$) has 1 atom. Carbon ($C$) has sub - script 2, so number of $C$ atoms is 2. Hydrogen ($H$) has sub - script 3, so number of $H$ atoms is 3. Oxygen ($O$) has sub - script 2, so number of $O$ atoms is 2. Total atoms = 1+2 + 3+2=8.

Step5: Analyze $NH_4Cl$

For $NH_4Cl$, nitrogen ($N$) has 1 atom. Hydrogen ($H$) has sub - script 4, so number of $H$ atoms is 4. Chlorine ($Cl$) has 1 atom. Total atoms = 1+4 + 1=6.

Step6: Analyze $HBr$

For $HBr$, hydrogen ($H$) has 1 atom and bromine ($Br$) has 1 atom. Total atoms = 1+1 = 2.

Step7: Analyze $4HNO_3$

The coefficient 4 multiplies each atom. Nitrogen ($N$) has 4 atoms. Hydrogen ($H$) has 4 atoms. Oxygen ($O$) has sub - script 3 and coefficient 4, so number of $O$ atoms is 4×3 = 12. Total atoms = 4+4+12 = 20.

Step8: Analyze $4Li_2O$

The coefficient 4 multiplies each atom. Lithium ($Li$) has sub - script 2 and coefficient 4, so number of $Li$ atoms is 4×2 = 8. Oxygen ($O$) has coefficient 4, so number of $O$ atoms is 4. Total atoms = 8+4 = 12.

Step9: Analyze $3H_2O$

The coefficient 3 multiplies each atom. Hydrogen ($H$) has sub - script 2 and coefficient 3, so number of $H$ atoms is 3×2 = 6. Oxygen ($O$) has coefficient 3, so number of $O$ atoms is 3. Total atoms = 6+3 = 9.

Step10: Analyze $3Al_2O_3$

The coefficient 3 multiplies each atom. Aluminum ($Al$) has sub - script 2 and coefficient 3, so number of $Al$ atoms is 3×2 = 6. Oxygen ($O$) has sub - script 3 and coefficient 3, so number of $O$ atoms is 3×3 = 9. Total atoms = 6+9 = 15.

Step11: Analyze $2CH_3OH$

The coefficient 2 multiplies each atom. Carbon ($C$) has 2 atoms. Hydrogen ($H$) has (3 + 1)×2=8 atoms. Oxygen ($O$) has 2 atoms. Total atoms = 2+8+2 = 12.

Step12: Analyze $2H_3NCH_3Br$

The coefficient 2 multiplies each atom. Hydrogen ($H$) has (3 + 3)×2 = 12 atoms. Nitrogen ($N$) has 2 atoms. Carbon ($C$) has 2 atoms. Bromine ($Br$) has 2 atoms. Total atoms = 12+2+2+2=18.

Answer:

1. $H_2O$: 3 atoms
2. $MgCl_2$: 3 atoms
3. $Li_2SO_4$: 7 atoms
4. $NaC_2H_3O_2$: 8 atoms
5. $NH_4Cl$: 6 atoms
6. $HBr$: 2 atoms
7. $4HNO_3$: 20 atoms
8. $4Li_2O$: 12 atoms
9. $3H_2O$: 9 atoms
10. $3Al_2O_3$: 15 atoms
11. $2CH_3OH$: 12 atoms
12. $2H_3NCH_3Br$: 18 atoms