Zn(CH3COO)2(H2O)2 (Zinc acetate) Gram to mole Conversion:
To convert gram weight to mole we need to find first what is the molecular weight of Zn(CH3COO)2(H2O)2 (Zinc acetate). Suppose the molecular weight of Zn(CH3COO)2(H2O)2 is M(Zn(CH3COO)2(H2O)2). Now we have to divide the gram weight by molecular mass of Zinc acetate. Then the result gives us a mole number.
mole number of Zn(CH3COO)2(H2O)2 (Zinc acetate) = {(grams weight of Zn(CH3COO)2(H2O)2) ÷ Molecular mass of Zn(CH3COO)2(H2O)2}
= {w÷ M(Zn(CH3COO)2(H2O)2)} [suppose w is grams weight of Zinc acetate ]
Example: Suppose a sample of Zn(CH3COO)2(H2O)2 has 185.76 grams. Now we have to calculate the mole number of the Zn(CH3COO)2(H2O)2.
Ans: mole = (184.93 ÷ 183.48 )
= 1.0079027686941 mole of Zinc acetate.
Zn(CH3COO)2(H2O)2 (Zinc acetate) grams to mole Conversion Table:
- 185.76 gram of Zinc acetate = 1.0124264224984 mole of Zinc acetate
- 188.67 gram of Zinc acetate = 1.0282864617397 mole of Zinc acetate
- 191.58 gram of Zinc acetate = 1.044146500981 mole of Zinc acetate
- 194.49 gram of Zinc acetate = 1.0600065402224 mole of Zinc acetate
- 197.4 gram of Zinc acetate = 1.0758665794637 mole of Zinc acetate
- 200.31 gram of Zinc acetate = 1.091726618705 mole of Zinc acetate
- 203.22 gram of Zinc acetate = 1.1075866579464 mole of Zinc acetate
- 206.13 gram of Zinc acetate = 1.1234466971877 mole of Zinc acetate
- 209.04 gram of Zinc acetate = 1.139306736429 mole of Zinc acetate
- 211.95 gram of Zinc acetate = 1.1551667756704 mole of Zinc acetate
- 214.86 gram of Zinc acetate = 1.1710268149117 mole of Zinc acetate
- 217.77 gram of Zinc acetate = 1.186886854153 mole of Zinc acetate
- 220.68 gram of Zinc acetate = 1.2027468933944 mole of Zinc acetate
- 223.59 gram of Zinc acetate = 1.2186069326357 mole of Zinc acetate
- 226.5 gram of Zinc acetate = 1.0758665794637 mole of Zinc acetate
- 229.41 gram of Zinc acetate = 1.2503270111184 mole of Zinc acetate
- 232.32 gram of Zinc acetate = 1.2661870503597 mole of Zinc acetate
- 235.23 gram of Zinc acetate = 1.282047089601 mole of Zinc acetate
- 238.14 gram of Zinc acetate = 1.2979071288424 mole of Zinc acetate
- 241.05 gram of Zinc acetate = 1.3137671680837 mole of Zinc acetate
- 243.96 gram of Zinc acetate = 1.329627207325 mole of Zinc acetate
- 246.87 gram of Zinc acetate = 1.3454872465664 mole of Zinc acetate
- 249.78 gram of Zinc acetate = 1.3613472858077 mole of Zinc acetate
- 252.69 gram of Zinc acetate = 1.3772073250491 mole of Zinc acetate
- 255.6 gram of Zinc acetate = 1.3930673642904 mole of Zinc acetate
- 258.51 gram of Zinc acetate = 1.4089274035317 mole of Zinc acetate
- 261.42 gram of Zinc acetate = 1.4247874427731 mole of Zinc acetate
- 264.33 gram of Zinc acetate = 1.4406474820144 mole of Zinc acetate
- 267.24 gram of Zinc acetate = 1.4565075212557 mole of Zinc acetate
- 270.15 gram of Zinc acetate = 1.4723675604971 mole of Zinc acetate
Zn(CH3COO)2(H2O)2 (Zinc acetate) mole to grams conversion:
If we know the mole number of Zinc acetate and we want to calculate grams then we again need molecular mass of Zn(CH3COO)2(H2O)2. We know that the molecular mass of Zn(CH3COO)2(H2O)2 is 183.48. To calculate grams we need to multiply mole numbers with molecular mass of Zinc acetate. If molecular mass of Zn(CH3COO)2(H2O)2 is M(Zn(CH3COO)2(H2O)2) and mole number is m(Zn(CH3COO)2(H2O)2) then
m(Zn(CH3COO)2(H2O)2) = (mole number of Zn(CH3COO)2(H2O)2 × molecular mass )
= {m(Zn(CH3COO)2(H2O)2) × M(Zn(CH3COO)2(H2O)2)}
Example: A sample of Zn(CH3COO)2(H2O)2 has 2.04 mole. We have to calculate how much grams of Zinc acetate it equivalent is.
Ans: Here mole number of Zn(CH3COO)2(H2O)2, m(Zn(CH3COO)2(H2O)2) = 2.04
Molecular weight of Zn(CH3COO)2(H2O)2, M(Zn(CH3COO)2(H2O)2) = 183.48
So, gram weight = ( 2.04 × 183.48 )
= 374.2992 gram
Zn(CH3COO)2(H2O)2 (Zinc acetate) mole to gram Conversion Table:
- 3.09 mole of Zinc acetate = 566.9532 gram of Zinc acetate
- 4.3 mole of Zinc acetate = 788.964 gram of Zinc acetate
- 5.51 mole of Zinc acetate = 1010.9748 gram of Zinc acetate
- 6.72 mole of Zinc acetate = 1232.9856 gram of Zinc acetate
- 7.93 mole of Zinc acetate = 1454.9964 gram of Zinc acetate
- 9.14 mole of Zinc acetate = 1677.0072 gram of Zinc acetate
- 10.35 mole of Zinc acetate = 1899.018 gram of Zinc acetate
- 11.56 mole of Zinc acetate = 2121.0288 gram of Zinc acetate
- 12.77 mole of Zinc acetate = 2343.0396 gram of Zinc acetate
- 13.98 mole of Zinc acetate = 2565.0504 gram of Zinc acetate
- 15.19 mole of Zinc acetate = 2787.0612 gram of Zinc acetate
- 16.4 mole of Zinc acetate = 3009.072 gram of Zinc acetate
- 17.61 mole of Zinc acetate = 3231.0828 gram of Zinc acetate
- 18.82 mole of Zinc acetate = 3453.0936 gram of Zinc acetate
- 20.03 mole of Zinc acetate = 3675.1044 gram of Zinc acetate
- 21.24 mole of Zinc acetate = 3897.1152 gram of Zinc acetate
- 22.45 mole of Zinc acetate = 4119.126 gram of Zinc acetate
- 23.66 mole of Zinc acetate = 4341.1368 gram of Zinc acetate
- 24.87 mole of Zinc acetate = 4563.1476 gram of Zinc acetate
- 26.08 mole of Zinc acetate = 4785.1584 gram of Zinc acetate
- 27.29 mole of Zinc acetate = 5007.1692 gram of Zinc acetate
- 28.5 mole of Zinc acetate = 5229.18 gram of Zinc acetate
- 29.71 mole of Zinc acetate = 5451.1908 gram of Zinc acetate
- 30.92 mole of Zinc acetate = 5673.2016 gram of Zinc acetate
- 32.13 mole of Zinc acetate = 5895.2124 gram of Zinc acetate
- 33.34 mole of Zinc acetate = 6117.2232 gram of Zinc acetate
- 34.55 mole of Zinc acetate = 6339.234 gram of Zinc acetate
- 35.76 mole of Zinc acetate = 6561.2448 gram of Zinc acetate
- 36.97 mole of Zinc acetate = 6783.2556 gram of Zinc acetate
- 38.18 mole of Zinc acetate = 7005.2664 gram of Zinc acetate
How to Calculate Molecular Mass of Zinc acetate (Zn(CH3COO)2(H2O)2):
For calculating molecular mass of Zinc acetate we need atomic mass of zinc, carbon, hydrogen and oxygen. We know that atomic masses of zinc (Zn), carbon (C), hydrogen (H) and oxygen (O) are 65.38, 12.011, 1.00797 and 15.9994 respectively. So molecular mass of Zinc acetate = [atomic mass of zinc (Zn) + {atomic mass of carbon (C) × 4} + {atomic mass of hydrogen (H) × 10} + {atomic mass of oxygen (O) × 6}]
= {65.38 + (12.011 × 4) + (1.00797 × 10) + (15.9994 × 6)}
= 219.5 gram