How many grams are present in 0.885 moles of manganese?
1 answer:
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374u
187u
C₁₄H₂₂N₄O₈
Explanation:
To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.
The empirical weight is determined by using the simplest ratio of the elements involved in the compound;
Molecular weight of C₁₄H₂₂N₄O₈;
atomic mass of C = 12g/mol
H = 1g/mol
N = 14g/mol
O = 16g/mol
Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)
= 168 + 22 + 56 + 128
= 374u
Empirical weight:
Empirical formula:
C₁₄ H₂₂ N₄ O₈
14 : 22 : 4 : 8
divide by 2:
7 : 11 : 2 : 4
empirical formula C₇H₁₁N₂O₄
empirical weight = =
= 187u
The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈
learn more:
Molecular mass brainly.com/question/5546238
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