1. Elementa masas daļu vielā aprēķina pēc formulas:

${w}_{\mathrm{el}.}=\frac{{m}_{\mathrm{el}.}}{{m}_{\mathrm{vielas}}}\cdot 100$

2. Ja ir $$1$$ vielas mols, tad tā masa ir vienāda ar

${m}_{\mathrm{vielas}}={n}_{\mathrm{vielas}}\cdot {M}_{\mathrm{vielas}}=1\phantom{\rule{0.147em}{0ex}}\mathrm{mol}\cdot M\phantom{\rule{0.147em}{0ex}}g/\mathrm{mol}$

3. $$1$$ vielas mols satur $$x$$ elementa molus, tāpēc vielas sastāvā esošā elementa masa ir vienāda ar

${m}_{\mathrm{el.}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{n}_{\mathrm{el.}}\cdot {M}_{\mathrm{el.}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\mathrm{mol}\cdot M\phantom{\rule{0.147em}{0ex}}g/\mathrm{mol}$

4. Ievietojot elementa masu un vielas masu formulā (1), iegūsim formulu (2), ko izmantosim aprēķiniem:

${w}_{\mathrm{el.}}=\frac{x\phantom{\rule{0.147em}{0ex}}\mathrm{mol}\phantom{\rule{0.147em}{0ex}}\left(\mathrm{el.}\right)\cdot M\phantom{\rule{0.147em}{0ex}}g/\mathrm{mol}\phantom{\rule{0.147em}{0ex}}\left(\mathrm{el.}\right)}{1\phantom{\rule{0.147em}{0ex}}\mathrm{mol}\cdot M\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}g/\mathrm{mol}\phantom{\rule{0.147em}{0ex}}\left(\mathrm{vielas}\right)}\cdot 100$

Piemērs:

Tādā veidā, piemēram, oglekļa masas daļas oglekļa(IV) oksīdā aprēķinam jāizskatās šādi:
1. No vielas formulas redzams, ka $$1$$ mols ${\mathrm{CO}}_{2}$ satur $$1$$ molu ķīmiskā elementa $C$.

2. Ja ir $$1$$ mols ${\mathrm{CO}}_{2}$, tad tā masa ir vienāda ar

${m}_{{\mathrm{CO}}_{2}}={n}_{{\mathrm{CO}}_{2}}\cdot {M}_{{\mathrm{CO}}_{2}}=1\phantom{\rule{0.147em}{0ex}}\mathrm{mol}\cdot 44\phantom{\rule{0.147em}{0ex}}g/\mathrm{mol}=44\phantom{\rule{0.147em}{0ex}}g$

3. Elementa $C$ $$1$$ molam masa ir

${m}_{C}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{n}_{C}\cdot {M}_{C}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}1\phantom{\rule{0.147em}{0ex}}\mathrm{mol}\cdot 12\phantom{\rule{0.147em}{0ex}}g/\mathrm{mol}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}12\phantom{\rule{0.147em}{0ex}}g$

4. Tātad elementa $C$ masas daļa vielā ${\mathrm{CO}}_{2}$ ir vienāda ar

${w}_{C}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{{m}_{C}}{{m}_{{\mathrm{CO}}_{2}}}\cdot 100\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{12\phantom{\rule{0.147em}{0ex}}g}{44\phantom{\rule{0.147em}{0ex}}g}\phantom{\rule{0.147em}{0ex}}\cdot 100\phantom{\rule{0.147em}{0ex}}\approx \phantom{\rule{0.147em}{0ex}}27,3%$

Atbilde: ${w}_{C}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}27,3%$

Materiālu izstrādāja A. Zaičenko