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counting atoms key vocabulary 1. a subscript = a number that is written…

Question

counting atoms
key vocabulary

  1. a subscript = 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.
  2. 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.
  3. 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.

Explanation:

Step1: Analyze \(H_2SO_4\)

  • For \(H\): The sub - script is \(2\), so \(H = 2\)
  • For \(S\): There is no sub - script, so \(S=1\)
  • For \(O\): The sub - script is \(4\), so \(O = 4\)
  • Total atoms \(=2 + 1+4=7\)

Step2: Analyze \(CH_3OH\)

  • For \(C\): There is no sub - script, so \(C = 1\)
  • For \(H\): \(3 + 1=4\) (from \(CH_3\) and \(OH\))
  • For \(O\): There is no sub - script, so \(O = 1\)
  • Total atoms \(=1+4 + 1=6\)

Step3: Analyze \(NaH_2PO_4\)

  • For \(Na\): There is no sub - script, so \(Na=1\)
  • For \(H\): The sub - script is \(2\), so \(H = 2\)
  • For \(P\): There is no sub - script, so \(P = 1\)
  • For \(O\): The sub - script is \(4\), so \(O = 4\)
  • Total atoms \(=1+2 + 1+4=8\)

Step4: Analyze \(NH_4Cl\)

  • For \(N\): There is no sub - script, so \(N = 1\)
  • For \(H\): The sub - script is \(4\), so \(H = 4\)
  • For \(Cl\): There is no sub - script, so \(Cl=1\)
  • Total atoms \(=1+4 + 1=6\)

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

  • For \(Al\): There is no sub - script, so \(Al = 1\)
  • For \(O\): \(1\times3 = 3\) (from \((OH)_3\))
  • For \(H\): \(1\times3 = 3\) (from \((OH)_3\))
  • Total atoms \(=1+3 + 3=7\)

Step6: Analyze \(Al_2(SO_4)_3\)

  • For \(Al\): The sub - script is \(2\), so \(Al = 2\)
  • For \(S\): \(1\times3 = 3\) (from \((SO_4)_3\))
  • For \(O\): \(4\times3 = 12\) (from \((SO_4)_3\))
  • Total atoms \(=2+3 + 12=17\)

Step7: Analyze \(2NaOH + H_2\)

  • For \(Na\): \(2\times1 = 2\) (from \(2NaOH\))
  • For \(O\): \(2\times1 = 2\) (from \(2NaOH\))
  • For \(H\): \(2\times1+2 = 4\) (from \(2NaOH\) and \(H_2\))
  • Total atoms \(=2+2 + 4=8\)

Step8: Analyze \(4Ca(HCO_3)_2\)

  • For \(Ca\): \(4\times1 = 4\)
  • For \(H\): \(1\times2\times4 = 8\) (from \((HCO_3)_2\) and coefficient \(4\))
  • For \(C\): \(1\times2\times4 = 8\) (from \((HCO_3)_2\) and coefficient \(4\))
  • For \(O\): \(3\times2\times4 = 24\) (from \((HCO_3)_2\) and coefficient \(4\))
  • Total atoms \(=4+8 + 8+24=44\)

Step9: Analyze \(4Li_2O\)

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

Step10: Analyze \(5ZnSO_4\)

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

Step11: Analyze \(Cu(NO_3)_2\)

  • For \(Cu\): There is no sub - script, so \(Cu = 1\)
  • For \(N\): \(1\times2 = 2\) (from \((NO_3)_2\))
  • For \(O\): \(3\times2 = 6\) (from \((NO_3)_2\))
  • Total atoms \(=1+2 + 6=9\)

Step12: Analyze \(3Pb(NO_3)_2\)

  • For \(Pb\): \(1\times3 = 3\)
  • For \(N\): \(1\times2\times3 = 6\) (from \((NO_3)_2\) and coefficient \(3\))
  • For \(O\): \(3\times2\times3 = 18\) (from \((NO_3)_2\) and coefficient \(3\))
  • 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. \(Al = 2\), \(S = 3\), \(O = 12\), total \(17\) (yellow)
  7. \(Na = 2\), \(O = 2\), \(H = 4\), total \(8\) (red)
  8. \(Ca = 4\), \(H = 8\), \(C = 8\), \(O = 24\), total \(44\) (sky blue)
  9. \(Li = 8\), \(O = 4\), total \(12\) (green)
  10. \(Zn = 5\), \(S = 5\), \(O = 20\), total \(30\) (red)
  11. \(Cu = 1\), \(N = 2\), \(O = 6\), total \(9\) (orange)
  12. \(Pb = 3\), \(N = 6\), \(O = 18\), total \(27\) (blue)