Ta có
n Na2CO3=\(\frac{1,272}{106}=0,012\left(mol\right)\)
n CO2=\(\frac{0,528}{44}=0,12\left(mol\right)\)
Ta có
n Na=2n Na2CO3=0,024(mol)
n C(Na2CO3)=n Na2CO3=0,012(mol)
n C(CO2)=n CO2=0,012(mol)
\(\sum n_C=0,012+0,012=0,024\left(mol\right)\)
m O=mA-mNa-m C=\(\text{1,608-0,024.23-0,024.12=0,768}\)
n O=\(\frac{0,768}{16}=0,048\left(mol\right)\)
\(n_C:n_O:n_{Na}=1:2:1\)
Do A chứa 2 nguyên tử A --->CTHH" C2O4Na2