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=(5×5 = 25)
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 1asbr₃ as = __ br = __
22 2h₂so₄ h = 4 s = 2 o = 8
23 5br₂ s = __ br = __
24 4ca(oh)₂ ca = 4 o = 8 h = 8
25 3mg₃(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 2krcl₂ kr = __ cl = __
31 5al(c₂h₃o₂)₃ 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 4au(no₃) au = __ n = o = __
38 3k₂zno₂ k = __ zn = o = __
39 3sr(mno₄)₂ sr = __ mn = o = __
40 4al(co₃)₃ al = __ c = o = __

Explanation:

Step1: Multiply coefficient by sub - script for each element

To find the number of atoms of an element in a given formula quantity, multiply the coefficient (the number in front of the formula) by the sub - script of the element. If there is no coefficient, it is 1, and if there is no sub - script, it is 1.
19. For \(8SnCl_{4}\):
- \(Sn\): Coefficient is 8, sub - script is 1. So number of \(Sn\) atoms \(=8\times1 = 8\)
- \(Cl\): Coefficient is 8, sub - script is 4. So number of \(Cl\) atoms \(=8\times4=32\)
21. For \(1AsBr_{3}\):
- \(As\): Coefficient is 1, sub - script is 1. So number of \(As\) atoms \(=1\times1 = 1\)
- \(Br\): Coefficient is 1, sub - script is 3. So number of \(Br\) atoms \(=1\times3 = 3\)
23. For \(5Br_{2}\):
- \(S\): There is no \(S\) in \(Br_{2}\), so number of \(S\) atoms \(=0\)
- \(Br\): Coefficient is 5, sub - script is 2. So number of \(Br\) atoms \(=5\times2 = 10\)
25. For \(5Mg_{3}(PO_{4})_{2}\):
- \(Mg\): Coefficient is 5, sub - script is 3. So number of \(Mg\) atoms \(=5\times3=15\)
- \(P\): Coefficient is 5, and considering the sub - script within the parentheses, \(P\) has sub - script 1 in \(PO_{4}\). So number of \(P\) atoms \(=5\times1\times2 = 10\)
- \(O\): Coefficient is 5, and \(O\) has sub - script 4 in \(PO_{4}\). So number of \(O\) atoms \(=5\times4\times2=40\)
27. For \(5N_{2}O_{4}\):
- \(N\): Coefficient is 5, sub - script is 2. So number of \(N\) atoms \(=5\times2 = 10\)
- \(O\): Coefficient is 5, sub - script is 4. So number of \(O\) atoms \(=5\times4 = 20\)
29. For \(7P_{2}O_{5}\):
- \(P\): Coefficient is 7, sub - script is 2. So number of \(P\) atoms \(=7\times2 = 14\)
- \(O\): Coefficient is 7, sub - script is 5. So number of \(O\) atoms \(=7\times5 = 35\)
30. For \(2KrCl_{2}\):
- \(Kr\): Coefficient is 2, sub - script is 1. So number of \(Kr\) atoms \(=2\times1 = 2\)
- \(Cl\): Coefficient is 2, sub - script is 2. So number of \(Cl\) atoms \(=2\times2 = 4\)
31. For \(5Al(C_{2}H_{3}O_{2})_{3}\):
- \(Al\): Coefficient is 5, sub - script is 1. So number of \(Al\) atoms \(=5\times1 = 5\)
- \(C\): Coefficient is 5, and \(C\) has sub - script 2 in \(C_{2}H_{3}O_{2}\). So number of \(C\) atoms \(=5\times2\times3 = 30\)
- \(H\): Coefficient is 5, and \(H\) has sub - script 3 in \(C_{2}H_{3}O_{2}\). So number of \(H\) atoms \(=5\times3\times3 = 45\)
- \(O\): Coefficient is 5, and \(O\) has sub - script 2 in \(C_{2}H_{3}O_{2}\). So number of \(O\) atoms \(=5\times2\times3 = 30\)
32. For \(3(NH_{4})_{2}Cr_{2}O_{7}\):
- \(N\): Coefficient is 3, and \(N\) has sub - script 1 in \(NH_{4}\). So number of \(N\) atoms \(=3\times1\times2 = 6\)
- \(H\): Coefficient is 3, and \(H\) has sub - script 4 in \(NH_{4}\). So number of \(H\) atoms \(=3\times4\times2 = 24\)
- \(Cr\): Coefficient is 3, and \(Cr\) has sub - script 2 in \((NH_{4})_{2}Cr_{2}O_{7}\). So number of \(Cr\) atoms \(=3\times2 = 6\)
- \(O\): Coefficient is 3, and \(O\) has sub - script 7 in \((NH_{4})_{2}Cr_{2}O_{7}\). So number of \(O\) atoms \(=3\times7 = 21\)
33. For \(5Fe_{2}(PO_{4})_{3}\):
- \(Fe\): Coefficient is 5, sub - script is 2. So number of \(Fe\) atoms \(=5\times2 = 10\)
- \(P\): Coefficient is 5, and \(P\) has sub - script 1 in \(PO_{4}\). So number of \(P\) atoms \(=5\times1\times3 = 15\)
- \(O\): Coefficient is 5, and \(O\) has sub - script 4 in \(PO_{4}\). So number of \(O\) atoms \(=5\times4\times3 = 60\)
34. For \(2NH_{4}NO_{3}\):
- \(N\): Coefficient is 2, and \(N\) has sub - script 1 in \(NH_{4}\) and 1 in \(NO_{3}\). So number of \(N\) atoms \(=2\times(1 + 1)=4\)
- \(H\): Coefficient is 2, and \(H\) has sub - script 4 in \(NH_{4}\). So number of \(H\) atoms \(=2\times4 = 8\)
- \(O\): Coefficient is 2, and \(O\) has sub - script 3 in \(NO_{3}\). So number of \(O\) atoms \(=2\times3 = 6\)
35. For \(5BaC_{4}H_{4}O_{6}\):
- \(Ba\): Coefficient is 5, sub - script is 1. So number of \(Ba\) atoms \(=5\times1 = 5\)
- \(C\): Coefficient is 5, sub - script is 4. So number of \(C\) atoms \(=5\times4 = 20\)
- \(H\): Coefficient is 5, sub - script is 4. So number of \(H\) atoms \(=5\times4 = 20\)
- \(O\): Coefficient is 5, sub - script is 6. So number of \(O\) atoms \(=5\times6 = 30\)
36. For \(4Cu(HSO_{4})_{2}\):
- \(Cu\): Coefficient is 4, sub - script is 1. So number of \(Cu\) atoms \(=4\times1 = 4\)
- \(H\): Coefficient is 4, and \(H\) has sub - script 1 in \(HSO_{4}\). So number of \(H\) atoms \(=4\times1\times2 = 8\)
- \(S\): Coefficient is 4, and \(S\) has sub - script 1 in \(HSO_{4}\). So number of \(S\) atoms \(=4\times1\times2 = 8\)
- \(O\): Coefficient is 4, and \(O\) has sub - script 4 in \(HSO_{4}\). So number of \(O\) atoms \(=4\times4\times2 = 32\)
37. For \(4Au(NO_{3})_{3}\):
- \(Au\): Coefficient is 4, sub - script is 1. So number of \(Au\) atoms \(=4\times1 = 4\)
- \(N\): Coefficient is 4, and \(N\) has sub - script 1 in \(NO_{3}\). So number of \(N\) atoms \(=4\times1\times3 = 12\)
- \(O\): Coefficient is 4, and \(O\) has sub - script 3 in \(NO_{3}\). So number of \(O\) atoms \(=4\times3\times3 = 36\)
38. For \(3K_{2}ZnO_{2}\):
- \(K\): Coefficient is 3, sub - script is 2. So number of \(K\) atoms \(=3\times2 = 6\)
- \(Zn\): Coefficient is 3, sub - script is 1. So number of \(Zn\) atoms \(=3\times1 = 3\)
- \(O\): Coefficient is 3, sub - script is 2. So number of \(O\) atoms \(=3\times2 = 6\)
39. For \(3Sr(MnO_{4})_{2}\):
- \(Sr\): Coefficient is 3, sub - script is 1. So number of \(Sr\) atoms \(=3\times1 = 3\)
- \(Mn\): Coefficient is 3, and \(Mn\) has sub - script 1 in \(MnO_{4}\). So number of \(Mn\) atoms \(=3\times1\times2 = 6\)
- \(O\): Coefficient is 3, and \(O\) has sub - script 4 in \(MnO_{4}\). So number of \(O\) atoms \(=3\times4\times2 = 24\)
40. For \(4Al(CO_{3})_{3}\):
- \(Al\): Coefficient is 4, sub - script is 1. So number of \(Al\) atoms \(=4\times1 = 4\)
- \(C\): Coefficient is 4, and \(C\) has sub - script 1 in \(CO_{3}\). So number of \(C\) atoms \(=4\times1\times3 = 12\)
- \(O\): Coefficient is 4, and \(O\) has sub - script 3 in \(CO_{3}\). So number of \(O\) atoms \(=4\times3\times3 = 36\)

Answer:

1. \(Sn = 8\), \(Cl = 32\)
2. \(As = 1\), \(Br = 3\)
3. \(S = 0\), \(Br = 10\)
4. \(Mg = 15\), \(P = 10\), \(O = 40\)
5. \(N = 10\), \(O = 20\)
6. \(P = 14\), \(O = 35\)
7. \(Kr = 2\), \(Cl = 4\)
8. \(Al = 5\), \(C = 30\), \(H = 45\), \(O = 30\)
9. \(N = 6\), \(H = 24\), \(Cr = 6\), \(O = 21\)
10. \(Fe = 10\), \(P = 15\), \(O = 60\)
11. \(N = 4\), \(H = 8\), \(O = 6\)
12. \(Ba = 5\), \(C = 20\), \(H = 20\), \(O = 30\)
13. \(Cu = 4\), \(H = 8\), \(S = 8\), \(O = 32\)
14. \(Au = 4\), \(N = 12\), \(O = 36\)
15. \(K = 6\), \(Zn = 3\), \(O = 6\)
16. \(Sr = 3\), \(Mn = 6\), \(O = 24\)
17. \(Al = 4\), \(C = 12\), \(O = 36\)