Ba(NO3)2 (Barium nitrate) Gram to mole Conversion:
To convert gram weight to mole we need to find first what is the molecular weight of Ba(NO3)2 (Barium nitrate). Suppose the molecular weight of Ba(NO3)2 is M(Ba(NO3)2). Now we have to divide the gram weight by molecular mass of Barium nitrate. Then the result gives us a mole number.
mole number of Ba(NO3)2 (Barium nitrate) = {(grams weight of Ba(NO3)2) ÷ Molecular mass of Ba(NO3)2}
= {w÷ M(Ba(NO3)2)} [suppose w is grams weight of Barium nitrate ]
Example: Suppose a sample of Ba(NO3)2 has 170.21 grams. Now we have to calculate the mole number of the Ba(NO3)2.
Ans: mole = (262.021 ÷ 261.337 )
= 1.0026173102163 mole of Barium nitrate.
Ba(NO3)2 (Barium nitrate) grams to mole Conversion Table:
- 170.21 gram of Barium nitrate = 0.65130463730738 mole of Barium nitrate
- 172.29 gram of Barium nitrate = 0.6592637093102 mole of Barium nitrate
- 174.37 gram of Barium nitrate = 0.66722278131302 mole of Barium nitrate
- 176.45 gram of Barium nitrate = 0.67518185331583 mole of Barium nitrate
- 178.53 gram of Barium nitrate = 0.68314092531865 mole of Barium nitrate
- 180.61 gram of Barium nitrate = 0.69109999732147 mole of Barium nitrate
- 182.69 gram of Barium nitrate = 0.69905906932428 mole of Barium nitrate
- 184.77 gram of Barium nitrate = 0.7070181413271 mole of Barium nitrate
- 186.85 gram of Barium nitrate = 0.71497721332992 mole of Barium nitrate
- 188.93 gram of Barium nitrate = 0.72293628533273 mole of Barium nitrate
- 191.01 gram of Barium nitrate = 0.73089535733555 mole of Barium nitrate
- 193.09 gram of Barium nitrate = 0.73885442933836 mole of Barium nitrate
- 195.17 gram of Barium nitrate = 0.74681350134118 mole of Barium nitrate
- 197.25 gram of Barium nitrate = 0.754772573344 mole of Barium nitrate
- 199.33 gram of Barium nitrate = 0.68314092531865 mole of Barium nitrate
- 201.41 gram of Barium nitrate = 0.77069071734963 mole of Barium nitrate
- 203.49 gram of Barium nitrate = 0.77864978935245 mole of Barium nitrate
- 205.57 gram of Barium nitrate = 0.78660886135526 mole of Barium nitrate
- 207.65 gram of Barium nitrate = 0.79456793335808 mole of Barium nitrate
- 209.73 gram of Barium nitrate = 0.8025270053609 mole of Barium nitrate
- 211.81 gram of Barium nitrate = 0.81048607736371 mole of Barium nitrate
- 213.89 gram of Barium nitrate = 0.81844514936653 mole of Barium nitrate
- 215.97 gram of Barium nitrate = 0.82640422136934 mole of Barium nitrate
- 218.05 gram of Barium nitrate = 0.83436329337216 mole of Barium nitrate
- 220.13 gram of Barium nitrate = 0.84232236537498 mole of Barium nitrate
- 222.21 gram of Barium nitrate = 0.85028143737779 mole of Barium nitrate
- 224.29 gram of Barium nitrate = 0.85824050938061 mole of Barium nitrate
- 226.37 gram of Barium nitrate = 0.86619958138343 mole of Barium nitrate
- 228.45 gram of Barium nitrate = 0.87415865338624 mole of Barium nitrate
- 230.53 gram of Barium nitrate = 0.88211772538906 mole of Barium nitrate
Ba(NO3)2 (Barium nitrate) mole to grams conversion:
If we know the mole number of Barium nitrate and we want to calculate grams then we again need molecular mass of Ba(NO3)2. We know that the molecular mass of Ba(NO3)2 is 261.337. To calculate grams we need to multiply mole numbers with molecular mass of Barium nitrate. If molecular mass of Ba(NO3)2 is M(Ba(NO3)2) and mole number is m(Ba(NO3)2) then
m(Ba(NO3)2) = (mole number of Ba(NO3)2 × molecular mass )
= {m(Ba(NO3)2) × M(Ba(NO3)2)}
Example: A sample of Ba(NO3)2 has 2.01 mole. We have to calculate how much grams of Barium nitrate it equivalent is.
Ans: Here mole number of Ba(NO3)2, m(Ba(NO3)2) = 2.01
Molecular weight of Ba(NO3)2, M(Ba(NO3)2) = 261.337
So, gram weight = ( 2.01 × 261.337 )
= 525.28737 gram
Ba(NO3)2 (Barium nitrate) mole to gram Conversion Table:
- 2.16 mole of Barium nitrate = 564.48792 gram of Barium nitrate
- 3.23 mole of Barium nitrate = 844.11851 gram of Barium nitrate
- 4.3 mole of Barium nitrate = 1123.7491 gram of Barium nitrate
- 5.37 mole of Barium nitrate = 1403.37969 gram of Barium nitrate
- 6.44 mole of Barium nitrate = 1683.01028 gram of Barium nitrate
- 7.51 mole of Barium nitrate = 1962.64087 gram of Barium nitrate
- 8.58 mole of Barium nitrate = 2242.27146 gram of Barium nitrate
- 9.65 mole of Barium nitrate = 2521.90205 gram of Barium nitrate
- 10.72 mole of Barium nitrate = 2801.53264 gram of Barium nitrate
- 11.79 mole of Barium nitrate = 3081.16323 gram of Barium nitrate
- 12.86 mole of Barium nitrate = 3360.79382 gram of Barium nitrate
- 13.93 mole of Barium nitrate = 3640.42441 gram of Barium nitrate
- 15 mole of Barium nitrate = 3920.055 gram of Barium nitrate
- 16.07 mole of Barium nitrate = 4199.68559 gram of Barium nitrate
- 17.14 mole of Barium nitrate = 4479.31618 gram of Barium nitrate
- 18.21 mole of Barium nitrate = 4758.94677 gram of Barium nitrate
- 19.28 mole of Barium nitrate = 5038.57736 gram of Barium nitrate
- 20.35 mole of Barium nitrate = 5318.20795 gram of Barium nitrate
- 21.42 mole of Barium nitrate = 5597.83854 gram of Barium nitrate
- 22.49 mole of Barium nitrate = 5877.46913 gram of Barium nitrate
- 23.56 mole of Barium nitrate = 6157.09972 gram of Barium nitrate
- 24.63 mole of Barium nitrate = 6436.73031 gram of Barium nitrate
- 25.7 mole of Barium nitrate = 6716.3609 gram of Barium nitrate
- 26.77 mole of Barium nitrate = 6995.99149 gram of Barium nitrate
- 27.84 mole of Barium nitrate = 7275.62208 gram of Barium nitrate
- 28.91 mole of Barium nitrate = 7555.25267 gram of Barium nitrate
- 29.98 mole of Barium nitrate = 7834.88326 gram of Barium nitrate
- 31.05 mole of Barium nitrate = 8114.51385 gram of Barium nitrate
- 32.12 mole of Barium nitrate = 8394.14444 gram of Barium nitrate
- 33.19 mole of Barium nitrate = 8673.77503 gram of Barium nitrate
How to Calculate Molecular Mass of Barium nitrate (Ba(NO3)2):
For calculating molecular mass of Barium nitrate we need atomic mass of barium, nitrogen and oxygen. We know that atomic masses of barium (Ba), nitrogen (N) and oxygen (O) are 137.33, 14.0067 and 15.9994 respectively. So molecular mass of Barium nitrate = [atomic mass of barium (Ba) + {atomic mass of nitrogen (N) + atomic mass of oxygen (O) × 3} × 2]
= [ 137.33 + {14.0067 + (15.9994 × 3)} × 2]
= (137.33 + 124.0098)
= 261.3398 gram