Question

instructions: write the quantity of atoms of each element opposite the formula of the compound for the quantity of formula units and molecules shown: for example: 5p₂o₅ p = (5×2 = )10 o = (5×3 = )15 for example: 4zn(no₃)₂ zn=(4×1 = )4 n=(4×1×2 = )8 o=(4×3×2 = )24 19. 8sncl₄ sn = __ cl = 20. 6cu₂seo₄ cu = 12 se = 6 o = 24 21. 3asbr₃ as = br = 22. 2h₂so₄ h = 4 s = 2 o = 8 23. sbr₂ s = br = 24. 4ca(oh)₂ ca = 4 o = 8 h = 8 25. 5mg₃(po₄)₂ mg = p = o = 26. 12h₂o h = 24 o = 12 27. 5n₂o₄ n = o = 28. 3clf cl = 3 f = 1 29. 7p₂o₅ p = o = 30. 2krcl₅ kr = cl = 31. 5al(c₂h₃o₂)₂ al = c = h = o = 32. 3(nh₄)₂cr₂o₇ n = h = cr = o = 33. 5fe₃(po₄)₂ fe = p = o = 34. 2nh₄no₃ n = h = o = 35. 5bac₄h₄o₆ ba = c = h = o = 36. 4cu(hso₃)₂ cu = h = s = o = 37. 9au(no₂)₂ au = n = o = 38. 3k₂zno₂ k = zn = o = 39. 3sr(mno₄)₂ sr = mn = o = 40. 4al₂(co₃)₃ al = c = o = __

Explanation:

Step1: Multiply coefficient by element sub - script

For each compound, multiply the coefficient (the number in front of the formula) by the sub - script of each element to find the number of atoms of that element. If there are parentheses, multiply the sub - script outside the parentheses by all sub - scripts inside.

1. For \(8SnCl_{4}\), \(Sn\): \(8\times1 = 8\), \(Cl\): \(8\times4=32\)
2. For \(3AsBr_{3}\), \(As\): \(3\times1 = 3\), \(Br\): \(3\times3 = 9\)
3. For \(SBr_{2}\), \(S\): \(1\times1 = 1\), \(Br\): \(1\times2 = 2\)
4. For \(5Mg_{3}(PO_{4})_{2}\), \(Mg\): \(5\times3=15\), \(P\): \(5\times1\times2 = 10\), \(O\): \(5\times4\times2=40\)
5. For \(5N_{2}O_{4}\), \(N\): \(5\times2 = 10\), \(O\): \(5\times4 = 20\)
6. For \(2KrCl_{5}\), \(Kr\): \(2\times1 = 2\), \(Cl\): \(2\times5 = 10\)
7. For \(5Al(C_{2}H_{3}O_{2})_{2}\), \(Al\): \(5\times1 = 5\), \(C\): \(5\times2\times2=20\), \(H\): \(5\times3\times2 = 30\), \(O\): \(5\times2\times2 = 20\)
8. For \(3(NH_{4})_{2}Cr_{2}O_{7}\), \(N\): \(3\times1\times2 = 6\), \(H\): \(3\times4\times2 = 24\), \(Cr\): \(3\times2 = 6\), \(O\): \(3\times7 = 21\)
9. For \(5Fe_{3}(PO_{4})_{2}\), \(Fe\): \(5\times3 = 15\), \(P\): \(5\times1\times2 = 10\), \(O\): \(5\times4\times2 = 40\)
10. For \(2NH_{4}NO_{3}\), \(N\): \(2\times2 = 4\), \(H\): \(2\times4 = 8\), \(O\): \(2\times3 = 6\)
11. For \(5BaC_{4}H_{4}O_{6}\), \(Ba\): \(5\times1 = 5\), \(C\): \(5\times4 = 20\), \(H\): \(5\times4 = 20\), \(O\): \(5\times6 = 30\)
12. For \(4Cu(HSO_{3})_{2}\), \(Cu\): \(4\times1 = 4\), \(H\): \(4\times1\times2 = 8\), \(S\): \(4\times1\times2 = 8\), \(O\): \(4\times3\times2 = 24\)
13. For \(9Au(NO_{2})_{2}\), \(Au\): \(9\times1 = 9\), \(N\): \(9\times1\times2 = 18\), \(O\): \(9\times2\times2 = 36\)
14. For \(3K_{2}ZnO_{2}\), \(K\): \(3\times2 = 6\), \(Zn\): \(3\times1 = 3\), \(O\): \(3\times2 = 6\)
15. For \(3Sr(MnO_{4})_{2}\), \(Sr\): \(3\times1 = 3\), \(Mn\): \(3\times1\times2 = 6\), \(O\): \(3\times4\times2 = 24\)
16. For \(4Al_{2}(CO_{3})_{3}\), \(Al\): \(4\times2 = 8\), \(C\): \(4\times1\times3 = 12\), \(O\): \(4\times3\times3 = 36\)

Answer:

1. \(Sn = 8\), \(Cl = 32\)
2. \(As = 3\), \(Br = 9\)
3. \(S = 1\), \(Br = 2\)
4. \(Mg = 15\), \(P = 10\), \(O = 40\)
5. \(N = 10\), \(O = 20\)
6. \(Kr = 2\), \(Cl = 10\)
7. \(Al = 5\), \(C = 20\), \(H = 30\), \(O = 20\)
8. \(N = 6\), \(H = 24\), \(Cr = 6\), \(O = 21\)
9. \(Fe = 15\), \(P = 10\), \(O = 40\)
10. \(N = 4\), \(H = 8\), \(O = 6\)
11. \(Ba = 5\), \(C = 20\), \(H = 20\), \(O = 30\)
12. \(Cu = 4\), \(H = 8\), \(S = 8\), \(O = 24\)
13. \(Au = 9\), \(N = 18\), \(O = 36\)
14. \(K = 6\), \(Zn = 3\), \(O = 6\)
15. \(Sr = 3\), \(Mn = 6\), \(O = 24\)
16. \(Al = 8\), \(C = 12\), \(O = 36\)