# What is the difference in Da of ESI analytes with multiple charges?

This is a follow-up question to Why are isotopes an issue in reading mass spectra?

A follow-up question: electrospray ionization (ESI) produces multiply charged species and charge can be >40 (for the purposes of m/z ratio), and assuming the difference in mass for an isotope is ~1 Da (with each additional neutron for an analyte at a charge of +1 in positive ionization mode) and assuming there is only one additional isotope, would the difference in weight for two isotopes each with a charge +10 be 0.1 Da?

would the difference in weight for two isotopes each with a charge +10 be 0.1 Da?

assuming there is only one additional isotope

This is a weird assumption, as it is often not true. Molecules containing both $$\ce{H}$$ atoms and $$\ce{C}$$ atoms will have isotopic contributions from $$\ce{H}$$, $$\ce{D}$$, $$\ce{^13C}$$, and $$\ce{^12C}$$. Anything that has a sulfur atom will have contributions from $$\ce{^32S}$$, $$\ce{^34S}$$, $$\ce{^33S}$$, and $$\ce{^36S}$$. However none of those things really affect the answer to your question as I understand it. Isotopes are separated by almost 1 (or 2) Da increments.

The key is almost. Ions with a +10 charge would have isotopologues that were almost 0.1 Da apart, but not exactly. If the ion in question were only made of carbon, then a more exact value would be 0.10034 Da apart. Not all instruments can resolve a mass difference of 0.1 Da from 0.10034 Da, but some can.

More realistically if you have a molecule with Br, then the distance between isotopes:

• Will be 2 $$m/z$$ units in case of 1 charge
• And 1 $$m/z$$ units in case of 2 charges

You can use this information to determine if there are molecules ionized twice — their isotopes appear closer than expected.

Eventually since ESI/APCI adds $$\ce{H+}$$ for each charge, given original EMW was 100 Da, then:

• $$m/z$$ in case of 1 charge: $$(100+1)/1=101$$ and $$(102+1)/1=103$$
• $$m/z$$ in case of 2 charges: $$(100+2)/2=51$$ and $$(102+2)/2=52$$

Or in case of negative ionization you'll need to subtract 1 and 2 Da, respectively.