Xác định tổng số nguyên tử trong Y.
Giải thích
Ta có: \(\% H = 100 - 32,3944 - 21,8310 - 45,0704 = 0,7042\)
CTTQ của Y dạng: NaxXaOyHz.
Ta có:\(\begin{array}{l}x:a:y:z = \frac{{32,3944}}{{23}}:\frac{{21,8310}}{{{M_X}}}:\frac{{45,0704}}{{16}}:\frac{{0,7042}}{1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\,\,\,1,41\,\,\,\,\,\;:\,\,\,\frac{{21,8310}}{{{M_X}}}\,\,:\,\,\,\,2,8169\,\,:\,\,\,0,7042\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,:\,\,\,\frac{{21,8310}}{{0,7042{M_X}}}\,:\,\,\,\,\,4\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,1\,\,\end{array}\)
Vậy: 21,83100,7042MX=a⇔31MX=a⇔a=1MX=31→CTTQ YNa2HPO4
Tổng số nguyên tử trong Y là 8