What's the empirical and molecular weight of C14H22N4O8? What's the molecular formula of C14H22N4O8?
1 answer:
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
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Molecular mass brainly.com/question/5546238
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Answer:
a)
b)
c)
Explanation:
a)
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Balancing of above reaction,
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(b) When aqueous is added to NaOH,
and
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Balancing of above reaction,
First balance all the atoms except O and H
S atom is already balanced in either side
No. of Na atom in left hand side = 1
No. of Na atom in right hand side = 2
So multiply NaOH by 2, now the reaction becomes:
No. of O atoms in left hand side = 6
No. of O atoms in right hand side = 5
So multiply H2O by 2, now the reaction becomes:
Now, it can been seen that all the atoms are balanced.
c) When reacts with HF, baroum fluoride and water is formed.
Balancing of above reaction,
Barium atoms are already balanced on either side.
No. of F atoms on left hand side = 1
No. of F atoms on right hand side = 2
So, multiply HF by 2, now reaction becomes
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