Answer:
- [tex]moles=0.0588mol\ Glucose[/tex]
- [tex]molecules=3.54x10^{22} molecules \ Glucose[/tex]
- [tex]atoms \ C=2.13x10^{23}atoms \ C[/tex]
Explanation:
Hello,
In this case, by knowing molar mass of glucose is 180.156 g/mol, we can apply the following mole-mass-particles relationships in order to compute the moles, molecules and carbon atoms in the 10.6-g sample of glucose:
- Moles:
[tex]moles=10.6g\ Glucose*\frac{1mol\ Glucose}{180.156g\ Glucose}\\\\moles=0.0588mol\ Glucose[/tex]
- Molecules: we also use Avogadro's number.
[tex]molecules=10.6g\ Glucose*\frac{1mol\ Glucose}{180.156g\ Glucose}*\frac{6.022x10^{23} molecules \ Glucose}{1mol\ Glucose} \\\\molecules=3.54x10^{22} molecules \ Glucose[/tex]
- Carbon atoms: here, particularly, one mole of glucose has six moles of carbon atoms, thus:
[tex]atoms\ C=10.6g\ Glucose*\frac{1mol\ Glucose}{180.156g\ Glucose}*\frac{6molC}{1mol\ Glucose} *\frac{6.022x10^{23}atoms \ C}{1molC} \\\\atoms \ C=2.13x10^{23}atoms \ C[/tex]
Regards.
